Journal of Inequalities in Pure and
Applied Mathematics
A SURVEY ON CAUCHY-BUNYAKOVSKY-SCHWARZ
TYPE DISCRETE INEQUALITIES
volume 4, issue 3, article 63,
2003.
S.S. DRAGOMIR
School of Computer Science and Mathematics
Victoria University of Technology
PO Box 14428
Melbourne City MC
8001 Victoria, Australia
EMail: sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 22 January, 2003;
accepted 14 May, 2003.
Communicated by: P.S. Bullen
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
010-03
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Abstract
The main purpose of this survey is to identify and highlight the discrete inequalities that are connected with (CBS)− inequality and provide refinements
and reverse results as well as to study some functional properties of certain
mappings that can be naturally associated with this inequality such as superadditivity, supermultiplicity, the strong versions of these and the corresponding
monotonicity properties. Many companion, reverse and related results both for
real and complex numbers are also presented.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
2000 Mathematics Subject Classification: 26D15, 26D10.
Key words: Cauchy-Bunyakovsky-Schwarz inequality, Discrete inequalities.
S.S. Dragomir
Contents
1
2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(CBS) – Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
(CBS) −Inequality for Real Numbers . . . . . . . . . . . .
2.2
(CBS) −Inequality for Complex Numbers . . . . . . . .
2.3
An Additive Generalisation . . . . . . . . . . . . . . . . . . . . .
2.4
A Related Additive Inequality . . . . . . . . . . . . . . . . . . .
2.5
A Parameter Additive Inequality . . . . . . . . . . . . . . . . .
2.6
A Generalisation Provided by Young’s Inequality . . .
2.7
Further Generalisations via Young’s Inequality . . . . .
2.8
A Generalisation Involving J−Convex Functions . . .
2.9
A Functional Generalisation . . . . . . . . . . . . . . . . . . . . .
2.10 A Generalisation for Power Series . . . . . . . . . . . . . . . .
2.11 A Generalisation of Callebaut’s Inequality . . . . . . . . .
6
9
9
11
13
17
19
22
24
34
36
40
43
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3
4
5
2.12 Wagner’s Inequality for Real Numbers . . . . . . . . . . . .
2.13 Wagner’s inequality for Complex Numbers . . . . . . . .
Refinements of the (CBS) −Inequality . . . . . . . . . . . . . . . . . .
3.1
A Refinement in Terms of Moduli . . . . . . . . . . . . . . . .
3.2
A Refinement for a Sequence Whose Norm is One . .
3.3
A Second Refinement in Terms of Moduli . . . . . . . . .
3.4
A Refinement for a Sequence Less than the Weights .
3.5
A Conditional Inequality Providing a Refinement . . .
3.6
A Refinement for Non-Constant Sequences . . . . . . . .
3.7
De Bruijn’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8
McLaughlin’s Inequality . . . . . . . . . . . . . . . . . . . . . . . .
3.9
A Refinement due to Daykin-Eliezer-Carlitz . . . . . . .
3.10 A Refinement via Dunkl-Williams’ Inequality . . . . . .
3.11 Some Refinements due to Alzer and Zheng . . . . . . . .
Functional Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
A Monotonicity Property . . . . . . . . . . . . . . . . . . . . . . .
4.2
A Superadditivity Property in Terms of Weights . . . .
4.3
The Superadditivity as an Index Set Mapping . . . . . .
4.4
Strong Superadditivity in Terms of Weights . . . . . . . .
4.5
Strong Superadditivity as an Index Set Mapping . . . .
4.6
Another Superadditivity Property . . . . . . . . . . . . . . . .
4.7
The Case of Index Set Mapping . . . . . . . . . . . . . . . . . .
4.8
Supermultiplicity in Terms of Weights . . . . . . . . . . . .
4.9
Supermultiplicity as an Index Set Mapping . . . . . . . .
Reverse Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
The Cassels’ Inequality . . . . . . . . . . . . . . . . . . . . . . . . .
5.2
The Pólya-Szegö Inequality . . . . . . . . . . . . . . . . . . . . .
46
49
55
55
59
63
67
72
76
83
84
86
89
91
104
104
106
109
112
116
119
124
128
134
143
143
147
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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6
5.3
The Greub-Rheinboldt Inequality . . . . . . . . . . . . . . . .
5.4
A Cassels’ Type Inequality for Complex Numbers . .
5.5
A Reverse Inequality for Real Numbers . . . . . . . . . . .
5.6
A Reverse Inequality for Complex Numbers . . . . . . .
5.7
Shisha-Mond Type Inequalities . . . . . . . . . . . . . . . . . .
5.8
Zagier Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . .
5.9
A Reverse Inequality in Terms of the sup −Norm . . .
5.10 A Reverse Inequality in Terms of the 1−Norm . . . . .
5.11 A Reverse Inequality in Terms of the p−Norm . . . . .
5.12 A Reverse Inequality Via an Andrica-Badea Result .
5.13 A Refinement of Cassels’ Inequality . . . . . . . . . . . . . .
5.14 Two Reverse Results Via Diaz-Metcalf Results . . . . .
5.15 Some Reverse Results Via the Čebyšev Functional . .
5.16 Another Reverse Result via a Grüss Type Result . . . .
Related Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Ostrowski’s Inequality for Real Sequences . . . . . . . . .
6.2
Ostrowski’s Inequality for Complex Sequences . . . . .
6.3
Another Ostrowski’s Inequality . . . . . . . . . . . . . . . . . .
6.4
Fan and Todd Inequalities . . . . . . . . . . . . . . . . . . . . . . .
6.5
Some Results for Asynchronous Sequences . . . . . . . .
6.6
An Inequality via A − G − H Mean Inequality . . . .
6.7
A Related Result via Jensen’s Inequality for Power
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.8
Inequalities Derived from the Double Sums Case . . .
6.9
A Functional Generalisation for Double Sums . . . . . .
6.10 A (CBS) −Type Result for Lipschitzian Functions .
6.11 An Inequality via Jensen’s Discrete Inequality . . . . . .
149
151
155
159
164
167
171
175
180
184
188
194
198
208
217
217
219
222
226
227
230
232
234
236
239
243
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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6.12 An Inequality via Lah-Ribarić Inequality . . . . . . . . . . 244
6.13 An Inequality via Dragomir-Ionescu Inequality . . . . . 247
6.14 An Inequality via a Refinement of Jensen’s Inequality250
6.15 Another Refinement via Jensen’s Inequality . . . . . . . . 254
6.16 An Inequality via Slater’s Result . . . . . . . . . . . . . . . . . 259
6.17 An Inequality via an Andrica-Raşa Result . . . . . . . . . 262
6.18 An Inequality via Jensen’s Result for Double Sums . 266
6.19 Some Inequalities for the Čebyšev Functional . . . . . . 269
6.20 Other Inequalities for the Čebyšev Functional . . . . . . 274
6.21 Bounds for the Čebyšev Functional . . . . . . . . . . . . . . . 277
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
References
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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1.
Introduction
The Cauchy-Bunyakovsky-Schwarz inequality, or for short, the (CBS)− inequality, plays an important role in different branches of Modern Mathematics including Hilbert Spaces Theory, Probability & Statistics, Classical Real
and Complex Analysis, Numerical Analysis, Qualitative Theory of Differential
Equations and their applications.
The main purpose of this survey is to identify and highlight the discrete inequalities that are connected with (CBS)− inequality and provide refinements
and reverse results as well as to study some functional properties of certain
mappings that can be naturally associated with this inequality such as superadditivity, supermultiplicity, the strong versions of these and the corresponding
monotonicity properties. Many companions and related results both for real and
complex numbers are also presented.
The first section is devoted to a number of (CBS)− type inequalities that
provides not only natural generalizations but also several extensions for different classes of analytic functions of a real variable. A generalization of the
Wagner inequality for complex numbers is obtained. Several results discovered
by the author in the late eighties and published in different journals of lesser
circulation are also surveyed.
The second section contains different refinements of the (CBS)− inequality
including de Bruijn’s inequality, McLaughlin’s inequality, the Daykin-EliezerCarlitz result in the version presented by Mitrinović-Pečarić and Fink as well as
the refinements of a particular version obtained by Alzer and Zheng. A number
of new results obtained by the author, which are connected with the above ones,
are also presented.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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Section 4 is devoted to the study of functional properties of different mappings naturally associated to the (CBS)− inequality. Properties such as superadditivity, strong superadditivity, monotonicity and supermultiplicity and the
corresponding inequalities are mentioned.
In the next section, Section 5, reverse results for the (CBS)− inequality are
surveyed. The results of Cassels, Pólya-Szegö, Greub-Rheinbold, Shisha-Mond
and Zagier are presented with their original proofs. New results and versions
for complex numbers are also obtained. Reverse results in terms of p−norms
of the forward difference recently discovered by the author and some refinements of Cassels and Pólya-Szegö results obtained via Andrica-Badea inequality are mentioned. Some new facts derived from Grüss type inequalities are also
pointed out.
Section 6 is devoted to various inequalities related to the (CBS)− inequality.
The two inequalities obtained by Ostrowski and Fan-Todd results are presented.
New inequalities obtained via Jensen type inequality for convex functions are
derived, some inequalities for the Čebyşev functionals are pointed out. Versions
for complex numbers that generalize Ostrowski results are also emphasised.
It was one of the main aims of the survey to provide complete proofs for the
results considered. We also note that in most cases only the original references
are mentioned. Each section concludes with a list of the references utilized and
thus may be read independently.
Being self contained, the survey may be used by both postgraduate students
and researchers interested in Theory of Inequalities & Applications as well as
by Mathematicians and other Scientists dealing with numerical computations,
bounds and estimates where the (CBS)− inequality may be used as a powerful
tool.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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The author intends to continue this survey with another one devoted to the
functional and integral versions of the (CBS)− inequality. The corresponding
results holding in inner-product and normed spaces will be considered as well.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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(CBS) – Type Inequalities
2.
(CBS) −Inequality for Real Numbers
2.1.
The following inequality is known in the literature as Cauchy’s or CauchySchwarz’s or Cauchy-Bunyakovsky-Schwarz’s inequality. For simplicity, we
shall refer to it throughout this work as the (CBS) −inequality.
Theorem 2.1. If ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) are sequences of real
numbers, then
!2
n
n
n
X
X
X
2
b2k
(2.1)
ak b k
≤
ak
k=1
k=1
1. Consider the quadratic polynomial P : R → R,
P (t) =
(2.2)
n
X
(ak t − bk )2 .
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It is obvious that
n
X
k=1
for any t ∈ R.
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k=1
P (t) =
S.S. Dragomir
k=1
with equality if and only if the sequences ā and b̄ are proportional, i.e., there is
a r ∈ R such that ak = rbk for each k ∈ {1, . . . , n} .
Proof.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
!
a2k
t2 − 2
n
X
k=1
!
ak b k
t+
n
X
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Since P (t) ≥ 0 for any t ∈ R it follows that the discriminant ∆ of P is
negative, i.e.,
!2
n
n
n
X
X
X
1
2
0≥ ∆=
ak
b2k
ak b k −
4
k=1
k=1
k=1
and the inequality (2.1) is proved.
2. If we use Lagrange’s identity
(2.3)
n
X
i=1
a2i
n
X
b2i
−
i=1
n
X
i=1
!2
ai b i
n
1X
=
(ai bj − aj bi )2
2 i,j=1
X
=
(ai bj − aj bi )2
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
1≤i<j≤n
then (2.1) obviously holds.
The equality holds in (2.1) iff
(ai bj − aj bi )2 = 0
for any i, j ∈ {1, . . . , n} which is equivalent with the fact that ā and b̄ are
proportional.
Remark 2.1. The inequality (2.1) apparently was firstly mentioned in the work
[2] of A.L. Cauchy in 1821. The integral form was obtained in 1859 by V.Y.
Bunyakovsky [1]. The corresponding version for inner-product spaces obtained
by H.A. Schwartz is mainly known as Schwarz’s inequality. For a short history
of this inequality see [3]. In what follows we use the spelling adopted in the
paper [3]. For other spellings of Bunyakovsky’s name, see MathSciNet.
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(CBS) −Inequality for Complex Numbers
2.2.
The following version of the (CBS) −inequality for complex numbers holds
[4, p. 84].
Theorem 2.2. If ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) are sequences of complex numbers, then
2
n
n
n
X
X
X
2
(2.4)
ak b k ≤
|ak |
|bk |2 ,
k=1
k=1
k=1
with equality if and only if there is a complex number c ∈ C such that ak = cb̄k
for any k ∈ {1, . . . , n} .
Proof.
1. For any complex number λ ∈ C one has the equality
(2.5)
Contents
k=1
=
ak − λb̄k
āk − λ̄bk
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J
k=1
=
n
X
k=1
S.S. Dragomir
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n
X
ak − λb̄k 2
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
n
X
X
2
2
2
|ak | + |λ|
|bk | − 2 Re λ̄
ak b k
k=1
If in (2.5) we choose λ0 ∈ C,
Pn
ak b k
λ0 := Pnk=1
2,
k=1 |bk |
k=1
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b̄ 6= 0
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then we get the identity
(2.6)
P
n
n
2
X
2 X
| nk=1 ak bk |
2
0≤
ak − λ0 b̄k =
|ak | − Pn
2 ,
k=1 |bk |
k=1
k=1
which proves (2.4).
By virtue of (2.6), we conclude that equality holds in (2.4) if and only if
ak = λ0 b̄k for any k ∈ {1, . . . , n} .
2. Using Binet-Cauchy’s identity for complex numbers
(2.7)
n
X
i=1
xi yi
n
X
i=1
zi ti −
n
X
i=1
xi ti
n
X
zi yi
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
i=1
n
1X
=
(xi zj − xj zi ) (yi tj − yj ti )
2 i,j=1
X
=
(xi zj − xj zi ) (yi tj − yj ti )
1≤i<j≤n
for the choices xi = āi , zi = bi , yi = ai , ti = b̄i , i = {1, . . . , n}, we get
2
n
n
n
n
X
X
X
1X
2
2
(2.8)
|ai |
|bi | − ai b i =
|āi bj − āj bi |2
2
i=1
i=1
i=1
i,j=1
X
=
|āi bj − āj bi |2 .
1≤i<j≤n
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Now the inequality (2.4) is a simple consequence of (2.8).
The case of equality is obvious by the identity (2.8) as well.
Remark 2.2. By the (CBS) −inequality for real numbers and the generalised
triangle inequality for complex numbers
n
n
X
X
|zi | ≥ zi , zi ∈ C, i ∈ {1, . . . , n}
i=1
i=1
we also have
n
2
X
ak b k ≤
n
X
!2
|ak bk |
k=1
k=1
≤
n
X
2
|ak |
k=1
n
X
|bk |2 .
S.S. Dragomir
k=1
Remark 2.3. The Lagrange identity for complex numbers stated in [4, p. 85] is
wrong. It should be corrected as in (2.8).
2.3.
An Additive Generalisation
The following generalisation of the (CBS) −inequality was obtained in [5, p.
5].
Theorem 2.3. If ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) , c̄ = (c1 , . . . , cn ) and
d̄ = (d1 , . . . , dn ) are sequences of real numbers and p̄ = (p1 , . . . , pn ) , q̄ =
(q1 , . . . , qn ) are nonnegative, then
(2.9)
n
X
i=1
pi a2i
n
X
i=1
qi b2i
+
n
X
i=1
pi c2i
n
X
i=1
qi d2i
≥2
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
i=1
p i ai c i
n
X
i=1
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qi bi di .
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If p̄ and q̄ are sequences of positive numbers, then the equality holds in (2.9) iff
ai bj = ci dj for any i, j ∈ {1, . . . , n} .
Proof. We will follow the proof from [5].
From the elementary inequality
(2.10)
a2 + b2 ≥ 2ab for any a, b ∈ R
with equality iff a = b, we have
(2.11)
a2i b2j + c2i d2j ≥ 2ai ci bj dj for any i, j ∈ {1, . . . , n} .
Multiplying (2.11) by pi qj ≥ 0, i, j ∈ {1, . . . , n} and summing over i and j
from 1 to n, we deduce (2.9).
If pi , qj > 0 (i = 1, . . . , n) , then the equality holds in (2.9) iff ai bj = ci dj
for any i, j ∈ {1, . . . , n} .
Remark 2.4. The condition ai bj = ci dj for ci 6= 0, bj 6= 0 (i, j = 1, . . . , n) is
d
equivalent with acii = bjj (i, j = 1, . . . , n) , i.e., ā, c̄ and b̄, d̄ are proportional
with the same constant k.
Remark 2.5. If in (2.9) we choose pi = qi = 1 (i = 1, . . . , n) , ci = bi , and
di = ai (i = 1, . . . , n) , then we recapture the (CBS) −inequality.
The following corollary holds [5, p. 6].
Corollary 2.4. If ā, b̄, c̄ and d̄ are nonnegative, then
" n
#
n
n
n
n
n
X
X
X
X
1 X 3 X 3
3
3
2 2
ai c i
bi di +
c i ai
(2.12)
di bi ≥
ai c i
b2i d2i ,
2 i=1
i=1
i=1
i=1
i=1
i=1
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
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#
" n
n
n
n
X
X
X
1 X 2
(2.13)
a bi di ·
b2i ai ci +
c2i bi di ·
d2i ai ci
2 i=1 i
i=1
i=1
i=1
≥
n
X
!2
ai bi ci di
.
i=1
Another result is embodied in the following corollary [5, p. 6].
Corollary 2.5. If ā, b̄, c̄ and d̄ are sequences of positive and real numbers,
then:
#
" n
n
n
n
n
n
X
X
X
1 X a3i X b3i X
(2.14)
+
ai c i
bi di ≥
a2i
b2i ,
2 i=1 ci i=1 di
i=1
i=1
i=1
i=1
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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"
(2.15)
#
n
n
n
n
X
1 X a2i bi X b2i ai X
+
bi c i
ai di ≥
2 i=1 ci i=1 di
i=1
i=1
n
X
!2
ai b i
Contents
.
i=1
Finally, we also have [5, p. 6].
Corollary 2.6. If ā, and b̄ are positive, then

!2 
n
n
n
n
n
3 X
3
X
X
X
X
ai
bi
1
2

−
ai b i
≥
ai
b2i −
2 i=1 bi i=1 ai
i=1
i=1
i=1
The following version for complex numbers also holds.
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X
i=1
!2
ai b i
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Theorem 2.7. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) , c̄ = (c1 , . . . , cn ) and
d̄ = (d1 , . . . , dn ) be sequences of complex numbers and p̄ = (p1 , . . . , pn ) ,
q̄ = (q1 , . . . , qn ) are nonnegative. Then one has the inequality
(2.16)
n
X
i=1
2
pi |ai |
n
X
2
qi |bi | +
i=1
n
X
i=1
2
pi |ci |
n
X
qi |di |2
i=1
"
≥ 2 Re
n
X
pi ai c̄i
i=1
n
X
#
qi bi d¯i .
i=1
The case of equality for p̄, q̄ positive holds iff ai bj = ci dj for any i, j ∈
{1, . . . , n} .
Proof. From the elementary inequality for complex numbers
|a|2 + |b|2 ≥ 2 Re ab̄ , a, b ∈ C,
with equality iff a = b, we have
(2.17)
¯
|ai |2 |bj |2 + |ci |2 |dj |2 ≥ 2 Re ai c̄i bj dj
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for any i, j ∈ {1, . . . , n} . Multiplying (2.17) by pi qj ≥ 0 and summing over i
and j from 1 to n, we deduce (2.16).
The case of equality is obvious and we omit the details.
Remark 2.6. Similar particular cases may be stated but we omit the details.
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2.4.
A Related Additive Inequality
The following inequality was obtained in [5, Theorem 1.1].
Theorem 2.8. If ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are sequences of real
numbers and c̄ = (c1 , . . . , cn ) , d̄ = (d1 , . . . , dn ) are nonnegative, then
(2.18)
n
X
i=1
di
n
X
ci a2i
+
n
X
i=1
i=1
ci
n
X
di b2i
i=1
≥2
n
X
c i ai
i=1
n
X
di bi .
i=1
If ci and di (i = 1, . . . , n) are positive, then equality holds in (2.18) iff ā = b̄ =
k̄ where k̄ = (k, k, . . . , k) is a constant sequence.
A Survey on
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Proof. We will follow the proof from [5].
From the elementary inequality
(2.19)
2
2
a + b ≥ 2ab for any a, b ∈ R
with equality iff a = b; we have
(2.20)
a2i + b2j ≥ 2ai bj for any i, j ∈ {1, . . . , n} .
Multiplying (2.20) by ci dj ≥ 0, i, j ∈ {1, . . . , n} and summing over i from 1 to
n and over j from 1 to n, we deduce (2.18).
If ci , dj > 0 (i = 1, . . . , n) , then the equality holds in (2.18) iff ai = bj for
any i, j ∈ {1, . . . , n} which is equivalent with the fact that ai = bi = k for any
i ∈ {1, . . . , n} .
The following corollary holds [5, p. 4].
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Corollary 2.9. If ā and b̄ are nonnegative sequences, then
" n
#
n
n
n
n
n
X
X
X
X
1 X 3X
3
2
(2.21)
a
bi +
ai
bi ≥
ai
b2i ;
2 i=1 i i=1
i=1
i=1
i=1
i=1
(2.22)
" n
#
n
n
n
X
X
1 X X 2
ai
a bi +
bi
b2i ai ≥
2 i=1 i=1 i
i=1
i=1
n
X
!2
ai b i
.
i=1
Another corollary that may be obtained is [5, p. 4 – 5].
Corollary 2.10. If ā and b̄ are sequences of positive real numbers, then
Pn 1 Pn 1
n
X
a2i + b2i
i=1 ai
i=1
Pn 1 bi ,
(2.23)
≥
2ai bi
i=1 ai bi
i=1
(2.24)
n
X
i=1
n
n
n
X
1 X 1 X
bi ≥ 2n2 ,
ai
+
b
a
i
i
i=1
i=1
i=1
and
(2.25)
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n
n
X
i=1
a2i + b2i
≥
2a2i b2i
n
X
i=1
1
ai
n
X
i=1
1
.
bi
The following version for complex numbers also holds.
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Theorem 2.11. If ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are sequences of complex
numbers, then for p̄ = (p1 , . . . , pn ) and q̄ = (q1 , . . . , qn ) two sequences of
nonnegative real numbers, one has the inequality
" n
#
n
n
n
n
n
X
X
X
X
X
X
qi
pi |ai |2 +
pi
qi |bi |2 ≥ 2 Re
p i ai
qi b̄i .
(2.26)
i=1
i=1
i=1
i=1
i=1
i=1
For p̄, q̄ positive sequences, the equality holds in (2.26) iff ā = b̄ = k̄ =
(k, . . . , k) .
The proof goes in a similar way with the one in Theorem 2.8 on making use
of the following elementary inequality holding for complex numbers
(2.27)
|a|2 + |b|2 ≥ 2 Re ab̄ , a, b ∈ C;
with equality iff a = b.
2.5.
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A Parameter Additive Inequality
The following inequality was obtained in [5, Theorem 4.1].
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Theorem 2.12. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be sequences of real
numbers and c̄ = (c1 , . . . , cn ) , d̄ = (d1 , . . . , dn ) be nonnegative. If α, β > 0
and γ ∈ R such that γ 2 ≤ αβ, then
(2.28)
α
n
X
i=1
di
n
X
i=1
a2i ci + β
n
X
i=1
ci
n
X
i=1
b2i di ≥ 2γ
n
X
i=1
c i ai
n
X
i=1
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Proof. We will follow the proof from [5].
Since α, β > 0 and γ 2 ≤ αβ, it follows that for any x, y ∈ R one has
αx2 + βy 2 ≥ 2γxy.
(2.29)
Choosing in (2.29) x = ai , y = bj (i, j = 1, . . . , n) , we get
αa2i + βb2j ≥ 2γai bj for any i, j ∈ {1, . . . , n} .
(2.30)
If we multiply (2.30) by ci dj ≥ 0 and sum over i and j from 1 to n, we deduce
the desired inequality (2.28).
S.S. Dragomir
The following corollary holds.
Corollary 2.13. If ā and b̄ are nonnegative sequences and α, β, γ are as in
Theorem 2.12, then
(2.31)
α
n
X
bi
i=1
(2.32)
α
n
X
i=1
n
X
a3i
+β
i=1
ai
n
X
i=1
n
X
ai
n
X
i=1
a2i bi
+β
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b3i
≥ 2γ
i=1
n
X
i=1
bi
n
X
n
X
a2i
i=1
b2i ai
≥ 2γ
i=1
The following particular case is important [5, p. 8].
n
X
Contents
b2i ,
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Theorem 2.14. Let ā, b̄ be sequences
of real numbers. If p̄ is a sequence of
Pn
nonnegative real numbers with i=1 pi > 0, then:
Pn
Pn
Pn
n
n
X
X
2
2
i=1 pi ai bi Pi=1 pi ai
i=1 pi bi
(2.33)
.
p i ai
p i bi ≥
n
i=1 pi
i=1
i=1
In particular,
(2.34)
n
X
a2i
i=1
n
n
X
b2i
i=1
n
n
X X
1X
ai b i
ai
bi .
≥
n i=1
i=1
i=1
Proof. We will follow the proof from [5, p. 8].
in Theorem 2.12,Pci = di = pi (i = 1, . . . , n) and α =
PnIf we 2choose P
n
n
2
p
b
,
β
=
i=1 i i
i=1 pi ai , γ =
i=1 pi ai bi , we observe, by the (CBS) −
inequality with the weights pi (i = 1, . . . , n) one has γ 2 ≤ αβ, and then by
(2.28) we deduce (2.33).
Remark 2.7. If we assume that ā and b̄ are asynchronous, i.e.,
(ai − aj ) (bi − bj ) ≤ 0 for any i, j ∈ {1, . . . , n} ,
then by Čebyšev’s inequality
(2.35)
n
X
p i ai
n
X
i=1
pi bi ≥
i=1
n
X
pi
n
X
i=1
(2.36)
i=1
ai
n
X
i=1
bi ≥ n
n
X
i=1
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p i ai b i
i=1
respectively
n
X
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we have the following refinements of the (CBS) −inequality
Pn
Pn
Pn
n
n
X
X
2
2
i=1 pi ai bi Pi=1 pi ai
i=1 pi bi
(2.37)
p i ai
p i bi ≥
n
i=1 pi
i=1
i=1
!
2
n
X
≥
p i ai b i
i=1
provided
Pn
i=1
pi ai bi ≥ 0, respectively
n
X
(2.38)
a2i
i=1
provided
n
X
i=1
n
n
n
X X
1X
b2i ≥
ai b i
ai
bi ≥
n i=1
i=1
i=1
n
X
!2
ai b i
i=1
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Pn
i=1 ai bi ≥ 0.
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2.6.
A Generalisation Provided by Young’s Inequality
The following result was obtained in [5, Theorem 5.1].
Theorem 2.15. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) , p̄ = (p1 , . . . , pn ) and
q̄ = (q1 , . . . , qn ) be sequences of nonnegative real numbers and α, β > 1 with
1
+ β1 = 1. Then one has the inequality
α
(2.39)
α
n
X
i=1
qi
n
X
i=1
pi bβi + β
n
X
i=1
pi
n
X
i=1
qi aαi ≥ αβ
n
X
i=1
p i bi
n
X
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q i ai .
i=1
If p̄ and q̄ are sequences of positive real numbers, then the equality holds in
(2.39) iff there exists a constant k ≥ 0 such that aαi = bβi = k for each i ∈
{1, . . . , n} .
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Proof. It is, by the Arithmetic-Geometric inequality [6, p. 15], well known that
(2.40)
1
1
1
1
1
1
x + y ≥ x α y β for x, y ≥ 0,
+ = 1, α, β > 1
α
β
α β
with equality iff x = y.
Applying (2.40) for x = aαi , y = bβj (i, j = 1, . . . , n) we have
(2.41)
αbβj + βaαi ≥ αβai bj for any i, j ∈ {1, . . . , n}
with equality iff aαi = bβj for any i, j ∈ {1, . . . , n} .
If we multiply (2.41) by qi pj ≥ 0 (i, j ∈ {1, . . . , n}) and sum over i and j
from 1 to n we deduce (2.39).
The case of equality is obvious by the above considerations.
The following corollary is a natural consequence of the above theorem.
Corollary 2.16. Let ā, b̄, α and β be as in Theorem 2.15. Then
(2.42)
(2.43)
n
n
n
n
n
n
1 X X α+1 1 X X β+1 X 2 X 2
bi
a
+
ai
b
≥
ai
bi ;
α i=1 i=1 i
β i=1 i=1 i
i=1
i=1
n
n
n
n
1X X α 1X X β
ai
b i ai +
bi
ai b i ≥
α i=1 i=1
β i=1 i=1
n
X
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The following result which provides a generalisation of the (CBS) −inequality
may be obtained by Theorem 2.15 as well [5, Theorem 5.2].
Theorem 2.17. Let x̄ and ȳ be sequences of positive real numbers. If α, β are
as above, then
! n
!2
n
n
n
X
X
X
X
1
1
(2.44)
xαi yi2−α +
xβi yi2−β ·
yi2 ≥
xi yi .
α i=1
β i=1
i=1
i=1
The equality holds iff x̄ and ȳ are proportional.
Proof. Follows by Theorem 2.15 on choosing pi = qi = yi2 , ai =
i ∈ {1, . . . , n} .
xi
,
yi
bi =
xi
,
yi
Remark 2.8. For α = β = 2, we recapture the (CBS) −inequality.
Remark 2.9. For ai = |zi | , bi = |wi | , with zi , wi ∈ C; i = 1, . . . , n, we may
obtain similar inequalities for complex numbers. We omit the details.
2.7.
Further Generalisations via Young’s Inequality
The following inequality is known in the literature as Young’s inequality
(2.45)
pxq + qy p ≥ pqxy,
x, y ≥ 0 and
1 1
+ = 1, p > 1
p q
with equality iff xq = y p .
The following result generalising the (CBS) −inequality was obtained in [7,
Theorem 2.1] (see also [8, Theorem 1]).
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Theorem 2.18. Let x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) be sequences of complex numbers and p̄ = (p1 , . . . , pn ) , q̄ = (q1 , . . . , qn ) be two sequences of
nonnegative real numbers. If p > 1, p1 + 1q = 1, then
(2.46)
n
n
n
n
X
X
1X
1X
pk |xk |p
qk |yk |p +
qk |xk |q
pk |yk |q
p k=1
q k=1
k=1
k=1
≥
n
X
pk |xk yk |
k=1
n
X
qk |xk yk | .
k=1
Proof. We shall follow the proof in [7].
Choosing x = |xj | |yi | , y = |xi | |yj | , i, j ∈ {1, . . . , n} , we get from (2.45)
(2.47)
q |xi |p |yj |p + p |xj |q |yi |q ≥ pq |xi yi | |xj yj |
for any i, j ∈ {1, . . . , n} .
Multiplying with pi qj ≥ 0 and summing over i and j from 1 to n, we deduce
the desired result (2.46).
The following corollary is a natural consequence of the above theorem [7,
Corollary 2.2] (see also [8, p. 105]).
Corollary 2.19. If x̄ and ȳ are as in Theorem 2.18 and m̄ = (m1 , . . . , mn ) is a
sequence of nonnegative real numbers, then
n
n
n
n
X
X
1X
1X
p
p
q
mk |xk |
mk |yk | +
mk |xk |
mk |yk |q
(2.48)
p k=1
q k=1
k=1
k=1
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≥
n
X
!2
mk |xk yk |
,
k=1
where p > 1,
1
p
+
1
q
= 1.
Remark 2.10. If in (2.48) we assume that mk = 1, k ∈ {1, . . . , n} , then we
obtain [7, p. 7] (see also [8, p. 105])
!2
n
n
n
n
n
X
X
X
1X
1X
p
p
q
q
|xk |
|yk | +
|xk |
|yk | ≥
|xk yk | ,
(2.49)
p k=1
q k=1
k=1
k=1
k=1
which, in the particular case p = q = 2 will provide the (CBS) −inequality.
The second generalisation of the (CBS) −inequality via Young’s inequality
is incorporated in the following theorem [7, Theorem 2.4] (see also [8, Theorem
2]).
Theorem 2.20. Let x̄, ȳ, p̄, q̄ and p, q be as in Theorem 2.18. Then one has the
inequality
(2.50)
≥
k=1
Proof. We shall follow the proof in [7].
p−1
pk |xk | |yk |
n
X
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n
n
n
n
X
X
1X
1X
pk |xk |p
qk |yk |q +
qk |xk |q
pk |yk |p
p k=1
q k=1
k=1
k=1
n
X
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qk |xk | |yk |q−1 .
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|x |
i|
Choosing in (2.45), x = |yjj | , y = |x
, we get
|yi |
q
p
|xj |
|xi |
|xi | |xj |
(2.51)
p
+q
≥ pq
|yj |
|yi |
|yi | |yj |
for any yi 6= 0, i, j ∈ {1, . . . , n} .
It is easy to see that (2.51) is equivalent to
(2.52)
q |xi |p |yj |q + p |yi |p |xj |q ≥ pq |xi | |yi |p−1 |xj | |yj |q−1
for any i, j ∈ {1, . . . , n} .
Multiplying (2.52) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and
j from 1 to n, we deduce the desired inequlality (2.50).
A Survey on
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The following corollary holds [7, Corollary 2.5] (see also [8, p. 106]).
Corollary 2.21. Let x̄, ȳ, m̄ and p̄, q̄ be as in Corollary 2.19. Then
n
n
n
n
X
X
1X
1X
(2.53)
mk |xk |p
mk |yk |q +
mk |xk |q
mk |yk |p
p k=1
q
k=1
k=1
k=1
≥
n
X
p−1
mk |xk | |yk |
k=1
n
X
mk |xk | |yk |q−1 .
k=1
Remark 2.11. If in (2.53) we assume that mk = 1, k ∈ {1, . . . , n} , then we
obtain [7, p. 8] (see also [8, p. 106])
n
n
n
n
X
X
1X
1X
p
q
q
|xk |
|yk | +
|xk |
|yk |p
(2.54)
p k=1
q k=1
k=1
k=1
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≥
n
X
|xk | |yk |p−1
n
X
|xk | |yk |q−1 ,
k=1
k=1
which, in the particular case p = q = 2 will provide the (CBS) −inequality.
The third result is embodied in the following theorem [7, Theorem 2.7] (see
also [8, Theorem 3]).
Theorem 2.22. Let x̄, ȳ, p̄, q̄ and p, q be as in Theorem 2.18. Then one has the
inequality
(2.55)
1
p
n
X
pk |xk |p
k=1
n
X
qk |yk |q +
k=1
1
q
n
X
qk |xk |p
k=1
≥
n
X
n
X
pk |yk |q
S.S. Dragomir
k=1
n
X
pk |xk yk |
k=1
pk |xk |p−1 |yk |q−1 .
k=1
JJ
J
(2.56)
q
p
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for any xi , yj 6= 0, i, j ∈ {1, . . . , n} , giving
q
Title Page
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Proof. We shall follow the proof in [7].
|yi |
|xi |
If we choose x = |y
and y = |x
in (2.45) we get
j|
j|
q
p
|yi |
|xi |
|xi | |yi |
p
+q
≥ pq
,
|yj |
|xj |
|xj | |yj |
p
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Close
p−1
q |xi | |yj | + p |yi | |xj | ≥ pq |xi yi | |xj |
q−1
|yj |
for any i, j ∈ {1, . . . , n} .
Multiplying (2.56) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and
j from 1 to n, we deduce the desired inequality (2.55).
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The following corollary is a natural consequence of the above theorem [8, p.
106].
Corollary 2.23. Let x̄, ȳ, m̄ and p̄, q̄ be as in Corollary 2.19. Then one has
the inequality:
(2.57)
n
X
p
mk |xk |
n
X
k=1
q
mk |yk | ≥
n
X
mk |xk yk |
mk |xk |p−1 |yk |q−1 .
k=1
k=1
k=1
n
X
Remark 2.12. If in (2.57) we assume that mk = 1, k = {1, . . . , n} , then we
obtain [7, p. 8] (see also [8, p. 10])
(2.58)
n
X
k=1
p
|xk |
n
X
k=1
q
|yk | ≥
n
X
|xk yk |
k=1
n
X
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S.S. Dragomir
|xk |p−1 |yk |q−1 ,
k=1
Title Page
which, in the particular case p = q = 2 will provide the (CBS) −inequality.
The fourth generalisation of the (CBS) −inequality is embodied in the following theorem [7, Theorem 2.9] (see also [8, Theorem 4]).
Theorem 2.24. Let x̄, ȳ, p̄, q̄ and p, q be as in Theorem 2.18. Then one has the
inequality
n
X
k=1
qk |xk yk |
n
X
k=1
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n
n
n
n
X
X
1X
1X
2
q
2
(2.59)
pk |xk |
qk |yk | +
pk |yk |
qk |xk |p
q k=1
p k=1
k=1
k=1
≥
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2
2
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pk |xk | q |yk | p .
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Proof. We shall follow the proof in [7].
2
2
Choosing in (2.45), x = |xi | q |yj | , y = |xj | |yi | p , we get
(2.60)
2
2
p |xi |2 |yj |q + q |xj |p |yi |2 ≥ pq |xi | q |yi | p |xj yj |
for any i, j ∈ {1, . . . , n} .
Multiply (2.60) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j
from 1 to n, we deduce the desired inequality (2.60).
The following corollary holds [7, Corollary 2.10] (see also [8, p. 107]).
Corollary 2.25. Let x̄, ȳ, m̄ and p, q be as in Corollary 2.19. Then one has the
inequality:
(2.61)
n
n
n
n
X
X
1X
1X
mk |xk |2
mk |yk |q +
mk |yk |2
mk |xk |p
q k=1
p
k=1
k=1
k=1
≥
n
X
mk |xk yk |
k=1
n
X
2
2
mk |xk | q |yk | p .
Remark 2.13. If in (2.61) we take mk = 1, k ∈ {1, . . . , n} , then we get
k=1
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n
n
n
n
X
X
1X
1X
2
q
2
(2.62)
|xk |
|yk | +
|yk |
|xk |p
q k=1
p
k=1
k=1
k=1
≥
S.S. Dragomir
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k=1
n
X
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|xk yk |
n
X
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2
q
2
p
|xk | |yk | ,
Page 30 of 288
k=1
which, in the particular case p = q = 2 will provide the (CBS) −inequality.
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The fifth result generalising the (CBS) −inequality is embodied in the following theorem [7, Theorem 2.12] (see also [8, Theorem 5]).
Theorem 2.26. Let x̄, ȳ, p̄, q̄ and p, q be as in Theorem 2.18. Then one has the
inequality
(2.63)
n
n
n
n
X
X
1X
1X
pk |xk |2
qk |yk |q +
pk |yk |2
qk |xk |p
p k=1
q
k=1
k=1
k=1
≥
n
X
2
p
pk |xk | |yk |
k=1
2
q
n
X
qk |xk |p−1 |yk |q−1 .
k=1
S.S. Dragomir
Proof. We will follow the proof in [7].
2
q
2
p
Choosing in (2.45), x = |y|yij| | , y = |x|xij| | , yi , xj 6= 0, i, j ∈ {1, . . . , n} , we
may write
!p
!q
2
2
2
2
p
q
|yi | q |xi | p
|xi |
|yi |
,
≥ pq
+q
p
|xj | |yj |
|xj |
|yj |
from where results
(2.64)
A Survey on
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2
2
p |yi |2 |xj |p + q |xi |2 |yj |q ≥ pq |xi | p |yi | q |xj |p−1 |yj |q−1
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for any i, j ∈ {1, . . . , n} .
Multiplying (2.64) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and
j from 1 to n, we deduce the desired inequality (2.63).
The following corollary holds [7, Corollary 2.13] (see also [8, p. 108]).
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Corollary 2.27. Let x̄, ȳ, m̄ and p, q be as in Corollary 2.19. Then one has the
inequality:
n
n
n
n
X
X
1X
1X
q
2
2
mk |yk | +
mk |xk |p
(2.65)
mk |xk |
mk |yk |
p k=1
q k=1
k=1
k=1
≥
n
X
2
2
mk |xk | p |yk | q
k=1
n
X
mk |xk |p−1 |yk |q−1 .
k=1
Remark 2.14. If in (2.46) we choose mk = 1, k ∈ {1, . . . , n} , then we get [7,
p. 10] (see also [8, p. 108])
S.S. Dragomir
n
n
n
n
X
X
1X
1X
|xk |2
|yk |q +
|yk |2
|xk |p
(2.66)
p k=1
q
k=1
k=1
k=1
≥
n
X
k=1
2
p
|xk | |yk |
A Survey on
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Type Discrete Inequalities
2
q
n
X
Title Page
|xk |p−1 |yk |q−1 ,
k=1
which in the particular case p = q = 2 will provide the (CBS) −inequality.
Finally, the following result generalising the (CBS) −inequality holds [7,
Theorem 2.15] (see also [8, Theorem 6]).
Theorem 2.28. Let x̄, ȳ, p̄, q̄ and p, q be as in Theorem 2.18. Then one has the
inequality:
n
n
n
n
X
X
1X
1X
2
p
2
(2.67)
pk |xk |
qk |yk | +
qk |yk |
pk |xk |q
p k=1
q k=1
k=1
k=1
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≥
n
X
2
pk |xk | p |yk |
k=1
n
X
2
qk |xk | q |yk | .
k=1
Proof. We shall follow the proof in [7].
From (2.45) one has the inequality
p
q
2
2
2
2
(2.68)
q |xi | p |yj | + p |xj | q |yi | ≥ pq |xi | p |yi | |xj | q |yj |
for any i, j ∈ {1, . . . , n} .
Multiplying (2.68) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and
j from 1 to n, we deduce the desired inequality (2.67).
The following corollary also holds [7, Corollary 2.16] (see also [8, p. 108]).
Corollary 2.29. With the assumptions in Corollary 2.19, one has the inequality
(2.69)
n
X
k=1
mk |xk |2
n
X
k=1
mk
1
1
|yk |p + |yk |q
p
q
≥
n
X
k=1
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S.S. Dragomir
Title Page
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2
mk |xk | p |yk |
n
X
2
mk |xk | q |yk | .
k=1
Remark 2.15. If in (2.69) we choose mk = 1 (k ∈ {1, . . . , n}) , then we get
X
n
n n
n
X
X
X
2
2
1
1
2
p
q
p
(2.70)
|xk |
|yk | + |yk | ≥
|xk | |yk |
|xk | q |yk | ,
p
q
k=1
k=1
k=1
k=1
which, in the particular case p = q = 2, provides the (CBS) −inequality.
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2.8.
A Generalisation Involving J−Convex Functions
For a > 1, we denote by expa the function
(2.71)
expa : R → (0, ∞) , expa (x) = ax .
Definition 2.1. A function f : I ⊆ R → R is said to be J−convex on an interval
I if
x+y
f (x) + f (y)
≤
for any x, y ∈ I.
(2.72)
f
2
2
It is obvious that any convex function on I is a J convex function on I, but
the converse does not generally hold.
The following lemma holds (see [7, Lemma 4.3]).
Lemma 2.30. Let f : I ⊆ R → R be a J−convex function on I, a > 1 and
x, y ∈ R\ {0} with loga x2 , loga y 2 ∈ I. Then loga |xy| ∈ I and
(2.73)
{expb [f (loga |xy|)]}2 ≤ expb f loga x2 expb f loga y 2
for any b > 1.
A Survey on
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Proof. I, being an interval, is a convex set in R and thus
1
loga |xy| =
loga x2 + loga y 2 ∈ I.
2
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Since f is J−convex, one has
1
2
2
f (loga |xy|) = f
loga x + loga y
2
f (loga x2 ) + f (loga y 2 )
≤
.
2
(2.74)
Taking the expb in both parts, we deduce
f (loga x2 ) + f (loga y 2 )
expb [f (loga |xy|)] ≤ expb
2
21
= expb f loga x2 expb f loga y 2
,
which is equivalent to (2.73).
Theorem 2.31. Let f : I ⊆ R → R be a J−convex function on I, a, b > 1
and ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) sequences of nonzero real numbers. If
loga a2k , loga b2k ∈ I for all k ∈ {1, . . . , n} , then one has the inequality:
(2.75)
S.S. Dragomir
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The following generalisation of the (CBS) −inequality in terms of a J−
convex function holds [7, Theorem 4.4].
( n
X
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)2
expb [f (loga |ak bk |)]
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k=1
≤
n
X
n
X
expb f loga a2k
expb f loga b2k .
k=1
k=1
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Proof. Using Lemma 2.30 and the (CBS) −inequality one has
n
X
expb [f (loga |ak bk |)]
k=1
≤
n
X
1
expb f loga a2k expb f loga b2k 2
k=1
≤
n n
n n
X
1 o2 X
1 o2
expb f loga a2k 2
expb f loga b2k 2
k=1
! 12
k=1
A Survey on
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Type Discrete Inequalities
which is clearly equivalent to (2.75).
S.S. Dragomir
Remark 2.16. If in (2.75) we choose a = b > 1 and f (x) = x, x ∈ R, then we
recapture the (CBS) −inequality.
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2.9.
A Functional Generalisation
The following result was proved in [10, Theorem 2].
Theorem 2.32. Let A be a subset of real numbers R, f : A → R and ā =
(a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) sequences of real numbers with the properties
that
(i) ai bi , a2i , b2i ∈ A for any i ∈ {1, . . . , n} ,
(ii) f (a2i ) , f (b2i ) ≥ 0 for any i ∈ {1, . . . , n} ,
(iii) f 2 (ai bi ) ≤ f (a2i ) f (b2i ) for any i ∈ {1, . . . , n} .
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Then one has the inequality:
" n
#2
n
n
X
X
X
2
(2.76)
f (ai bi ) ≤
f ai
f b2i .
i=1
i=1
i=1
Proof. We give here a simpler proof than that found in [10].
We have
n
n
X
X
f (ai bi ) ≤
|f (ai bi )|
i=1
≤
i=1
n
X
1
1 f a2i 2 f b2i 2
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
i=1
"
n n X
21 2
12 2 X
2
2
f bi
≤
f ai
i=1
"
=
n
X
i=1
# 12
(by the (CBS)-ineq.)
i=1
f a2i
n
X
# 21
f b2i
i=1
and the inequality (2.76) is proved.
Remark 2.17. It is obvious that for A = R and f (x) = x, we recapture the
(CBS) −inequality.
Assume that ϕ : N → N is Euler’s indicator. In 1940, T. Popoviciu [11]
proved the following inequality for ϕ
(2.77)
[ϕ (ab)]2 ≤ ϕ a2 ϕ b2 for any natural number a, b;
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with equality iff a and b have the same prime factors.
A simple proof of this fact may be done by using the representation
1
1
ϕ (n) = n 1 −
··· 1 −
,
p1
pk
where n = pα1 1 pα2 2 · · · pαk k [9, p. 109].
The following generalisation of Popoviciu’s result holds [10, Theorem 1].
Theorem 2.33. Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality
"
(2.78)
n
X
#2
≤
ϕ (ai bi )
i=1
n
X
ϕ
a2i
i=1
n
X
ϕ b2i .
Further, let us denote by s (n) the sum of all relatively prime numbers with
n and less than n. Then the following result also holds [10, Theorem 1].
Theorem 2.34. Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality
(2.79)
n
X
i=1
#2
s (ai bi )
≤
n
X
i=1
S.S. Dragomir
i=1
Proof. Follows by Theorem 2.32 on taking into account that, by (2.77),
[ϕ (ai bi )]2 ≤ ϕ a2i ϕ b2i for any i ∈ {1, . . . , n} .
"
A Survey on
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Type Discrete Inequalities
s
a2i
n
X
i=1
s b2i .
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Proof. It is known (see for example [9, p. 109]) that for any n ∈ N one has
1
s (n) = nϕ (n) .
2
(2.80)
Thus
(2.81)
1
1
[s (ai bi )]2 = a2i b2i ϕ2 (ai bi ) ≤ a2i b2i ϕ a2i ϕ b2i = s a2i s b2i
4
4
for each i ∈ {1, . . . , n} .
Using Theorem 2.32 we then deduce the desired inequality (2.79).
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
The following corollaries of Theorem 2.32 are also natural to be considered
[10, p. 126].
S.S. Dragomir
Corollary 2.35. Let ai , bi ∈ R (i = 1, . . . , n) and a > 1. Denote expa x = ax ,
x ∈ R. Then one has the inequality
" n
#2
n
n
X
X
X
2
(2.82)
expa (ai bi ) ≤
expa ai
expa b2i .
Title Page
i=1
i=1
i=1
Corollary 2.36. Let ai , bi ∈ (−1, 1) (i = 1, . . . , n) and m > 0. Then one has
the inequality:
#2
" n
n
n
X
X
X
1
1
1
≤
.
(2.83)
m
m
2
2 m
(1
−
a
b
)
(1
−
a
)
(1
−
b
)
i
i
i
i
i=1
i=1
i=1
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2.10.
A Generalisation for Power Series
The following result holds [12, Remark 2].
P
k
Theorem 2.37. Let F : (−r, r) → R, F (x) = ∞
k=0 αk x with αk ≥ 0, k ∈ N.
If ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are sequences of real numbers such that
(2.84)
ai bi , a2i , b2i ∈ (−r, r) for any i ∈ {1, . . . , n} ,
then one has the inequality:
n
X
(2.85)
F
a2i
i=1
n
X
"
F
b2i
≥
i=1
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
#2
F (ai bi )
.
Proof. Firstly, let us observe that if x, y ∈ R such that xy, x2 , y 2 ∈ (−r, r) ,
then one has the inequality
(2.86)
[F (xy)]2 ≤ F x2 F y 2 .
Indeed, by the (CBS) −inequality, we have
"
(2.87)
n
X
k=0
#2
k k
αk x y
≤
S.S. Dragomir
i=1
n
X
k=0
αk x
2k
n
X
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2k
αk y ,
k=0
Taking the limit as n → ∞ in (2.87), we deduce (2.86).
n ≥ 0.
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Using the (CBS) −inequality and (2.86) we have
n
n
X
X
F (ai bi ) ≤
|F (ai bi )|
i=1
≤
i=1
n
X
1
1 F a2i 2 F b2i 2
i=1
( n
)1
n 2
X 1 2 X
1 2
F b2i 2
≤
F a2i 2
i=1
"
=
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
i=1
n
X
2
F ai
i=1
# 12
2
S.S. Dragomir
,
F bi
i=1
which is clearly equivalent to (2.85).
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The following particular inequalities of (CBS) −type hold [12, p. 164].
1. If ā, b̄ are sequences of real numbers, then one has the inequality
" n
#2
n
n
X
X
X
(2.88)
exp a2k
exp b2k ≥
exp (ak bk ) ;
k=1
(2.89)
n
X
sinh
k=1
a2k
n
X
k=1
sinh
b2k
≥
k=1
k=1
(2.90)
n
X
k=1
"
cosh a2k
n
X
k=1
n
X
#2
sinh (ak bk )
#2
n
X
cosh b2k ≥
cosh (ak bk )
k=1
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k=1
"
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2. If ā, b̄ are such that ai , bi ∈ (−1, 1) , i ∈ {1, . . . , n} , then one has the
inequalities
" n
#2
n
n
X
X
X
(2.91)
tan a2k
tan b2k ≥
tan (ak bk ) ;
k=1
n
X
(2.92)
k=1
2
arcsin ak
(2.93)
ln
n Y
1 + a2
k=1
k=1
"
2
arcsin bk ≥
k=1
k=1
"
n
X
1−
k
a2k
#
n
X
#2
arcsin (ak bk )
k=1
"
n Y
1 + b2
k
ln
k=1
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
#
1 − b2k
( "
≥
ln
S.S. Dragomir
n Y
1 + ak b k
k=1
"
(2.94)
ln
n Y
k=1
1
1 − a2k
#
"
ln
n Y
k=1
#
1
1 − b2k
( "
≥
ln
n
X
k=1
"
#)2
1 − ak b k
;
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Contents
JJ
J
n Y
k=1
(2.95)
;
1
1 − ak b k
n
n
X
X
1
1
1
≥
2 m
2 m
(1 − ak bk )m
(1 − ak ) k=1 (1 − bk )
k=1
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#)2
;
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#2
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, m > 0.
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2.11.
A Generalisation of Callebaut’s Inequality
The following result holds (see also [12, Theorem 2] for a generalisation for
positive linear functionals).
P∞
k
Theorem 2.38. Let F : (−r, r) → R, F (x) =
k=0 αk x with αk ≥ 0,
k ∈ N. If ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are sequences of nonnegative real
numbers such that
(2.96)
ai bi , aαi b2−α
, a2−α
bαi ∈ (0, r) for any i ∈ {1, . . . , n} ; α ∈ [0, 2] ,
i
i
then one has the inequality
" n
#2
n
n
X
X
X
α 2−α
α
(2.97)
F (ai bi ) ≤
F ai b i
F a2−α
b
i .
i
i=1
i=1
Indeed, using Callebaut’s inequality, i.e., we recall it [4]
!2
m
m
m
X
X
X
α 2−α
(2.99)
αi xi yi
≤
αi xi yi
αi x2−α
yiα ,
i
i=1
i=0
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i=1
we may write, for m ≥ 0, that
!2
m
m
m
X
X
X
i
i i
α 2−α i
(2.100)
αi x y
≤
αi x y
αi x2−α y α .
i=0
S.S. Dragomir
i=1
Proof. Firstly, we note that for any x, y > 0 such that xy, xα y 2−α , x2−α y α ∈
(0, r) one has
(2.98)
[F (xy)]2 ≤ F xα y 2−α F x2−α y α .
i=1
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Type Discrete Inequalities
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Taking the limit as m → ∞, we deduce (2.98).
Using the (CBS) −inequality and (2.98) we may write:
n
n
X
X
F (ai bi ) ≤
|F (ai bi )|
i=1
≤
i=1
n
X
1
21 F aαi b2−α
F a2−α
bαi 2
i
i
i=1
)1
( n
n 2
X 1 2 X
1 2
2
≤
F aαi b2−α
F a2−α
bαi 2
i
i
i=1
"
=
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
i=1
n
X
2−α
F aαi bi
i=1
S.S. Dragomir
# 21
F a2−α
bi
i
α
Title Page
i=1
Contents
which is clearly equivalent to (2.97).
The following particular inequalities also hold [12, pp. 165-166].
1. Let ā and b̄ be sequences of nonnegative real numbers. Then one has the
inequalities
" n
#2
n
n
X
X
X
α 2−α
(2.101)
exp (ak bk ) ≤
exp ak bk
exp ak2−α bαk ;
k=1
"
(2.102)
n
X
k=1
k=1
#2
sinh (ak bk )
≤
n
X
k=1
k=1
sinh
aαk b2−α
k
n
X
k=1
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sinh
ak2−α bαk
;
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"
(2.103)
n
X
#2
≤
cosh (ak bk )
k=1
n
X
cosh aαk b2−α
k
n
X
cosh a2−α
bαk .
k
k=1
k=1
2. Let ā and b̄ be such that ak , bk ∈ (0, 1) for any k ∈ {1, . . . , n} . Then one
has the inequalities:
" n
#2
n
n
X
X
X
α 2−α
(2.104)
tan ak2−α bαk ;
tan (ak bk ) ≤
tan ak bk
k=1
"
(2.105)
n
X
k=1
k=1
#2
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
arcsin (ak bk )
k=1
≤
n
X
arcsin aαk b2−α
k
k=1
( "
(2.106)
ln
n Y
1 + ak b k
(2.107)
ln
arcsin ak2−α bαk ;
k=1
#)2
1 − ak b k
" n # " n #
Y 1 + a2−α bα Y 1 + aα b2−α k k
k
k
≤ ln
ln
;
2−α α
α 2−α
1
−
a
b
1
−
a
b
k
k
k
k
k=1
k=1
k=1
( "
n
X
n Y
k=1
1
1 − ak b k
#)2
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"
≤ ln
n Y
k=1
2.12.
1
1 − aαk b2−α
k
#
"
ln
n Y
k=1
1
1 − a2−α
bαk
k
#
.
Wagner’s Inequality for Real Numbers
The following generalisation of the (CBS) − inequality for sequences of real
numbers is known in the literature as Wagner’s inequality [15], or [14] (see also
[4, p. 85]).
Theorem 2.39. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of real
numbers. If 0 ≤ x ≤ 1, then one has the inequality
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
(2.108)
n
X
!2
X
ak b k + x
k=1
ai b j
Title Page
1≤i6=j≤n
"
≤
n
X
#"
a2k
X
+ 2x
ai aj
1≤i<j≤n
k=1
n
X
#
b2k + 2x
k=1
Proof. We shall follow the proof in [13] (see also [4, p. 85]).
For any x ∈ [0, 1] , consider the quadratic polynomial in y
" n
#2
n
X
X
P (y) := (1 − x)
(ak y − bk )2 + x
(ak y − bk )
k=1
"
= (1 − x) y
2
k=1
n
X
k=1
a2k
− 2y
n
X
k=1
ak b k +
n
X
k=1
#
b2k
X
1≤i<j≤n
bi bj .
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
!2
n
X
+ x y 2
− 2y
ak
k=1

= (1 − x)
n
X
!
n
X
ak
n
X
a2k + x
!
!2 
k=1
n
X
ak
k=1
n
X
n
X
n
X
bk + (1 − x)
k=1
b2k + x
k=1
n
X
k=1
+
b2k + x 
n
X
k=1
k=1
!2
bk
k=1
bk
−
k=1
n
X
k=1
S.S. Dragomir
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JJ
J
b2k  .
k=1
II
I
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n
X
k=1
A Survey on
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Type Discrete Inequalities

Since, it is obvious that:
n
X

!2
k=1

n
X
bk
k=1
#

!2 
ak b k

!2
n
n

X
X
2
2

ak −
=
ak + x
ak
y2


k=1
k=1
k=1
" n
!#
n
n
n
X
X
X
X
ak b k + x
ak
bk −
ak b k
− 2y

n
X
n
X
k=1
"
 y 2 − 2y (1 − x)
ak
+
bk
k=1
k=1
k=1
+x
n
X
!2
ak
k=1
n
X
ak
k=1
bk −
−
n
X
k=1
n
X
a2k = 2
ak b k =
k=1
Close
X
ai aj ,
1≤i<j≤n
X
1≤i6=j≤n
Quit
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ai b j
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and
n
X
!2
−
bk
k=1
n
X
X
b2k = 2
bi bj ,
1≤i<j≤n
k=1
we get
P (y) =
n
X
!
a2k
X
+ 2x
ai aj
y2
1≤i<j≤n
k=1
− 2y
n
X
!
X
ak b k + x
k=1
ai b j
1≤i6=j≤n
+
n
X
b2k + 2x
k=1
X
bi bj .
1≤i<j≤n
S.S. Dragomir
Taking into consideration, by the definition of P, that P (y) ≥ 0 for any y ∈ R,
it follows that the discriminant ∆ ≤ 0, i.e.,
1
0≥ ∆=
4
n
X
ak b k + x
k=1
−
ai b j
1≤i6=j≤n
n
X
k=1
!
X
a2k + 2x
1≤i<j≤n
ai aj
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!2
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
k=1
!
b2k + 2x
X
bi bj
1≤i<j≤n
and the inequality (2.108) is proved.
Remark 2.18. If x = 0, then from (2.108) we recapture the (CBS) −inequality
for real numbers.
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2.13.
Wagner’s inequality for Complex Numbers
The following inequality which provides a version for complex numbers of
Wagner’s result holds [16].
Theorem 2.40. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of
complex numbers. Then for any x ∈ [0, 1] one has the inequality
#2
" n
X
X
(2.109)
Re ak b̄k + x
Re ai b̄j
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
1≤i6=j≤n
k=1
"
≤
n
X
#
X
|ak |2 + 2x
Re (ai āj )
S.S. Dragomir
1≤i<j≤n
k=1
"
n
X
#
X
|bk |2 + 2x
Re bi b̄j
1≤i<j≤n
k=1
(2.110)
f (t) = (1 − x)
k=1
2
n
X
2
|tak − bk | + x (tak − bk ) .
k=1
We have
(2.111) f (t) = (1 − x)
n
X
.
Title Page
Contents
Proof. Start with the function f : R → R,
n
X
JJ
J
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(tak − bk ) tāk − b̄k
Quit
k=1
+x t
n
X
k=1
ak −
n
X
k=1
!
bk
t
n
X
k=1
āk −
n
X
k=1
!
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b̄k
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"
= (1 − x) t2
n
X
2
|ak | − t
k=1
"
2
+x t
n
X
2
|ak | − t
k=1
n
X
ak
= (1 − x)
k=1
n
X
n
X
n
X
bk
n
X
b̄k +
k=1
+ 2 (1 − x)
|bk |
k=1
āk
n
X
#
2
|bk |
k=1
2 
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
2
k=1
n
X
S.S. Dragomir
"
Re ak b̄k + x Re
n
X
k=1
2
n
n
X
X
+ (1 − x)
|bk |2 + x bk .
k=1
k=1
Observe that
i,j=1
n
X
2
|ai | +
i=1
ak
n
X
k=1
##
b̄k
t
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2
n
n
X
X
ak =
ai āj
=
#
2
k=1
k=1
k=1
n
X
n
X
|ak | + x ak  t2
"
(2.112)
ak b̄k +
k=1
k=1
k=1
n
X
bk āk − t
k=1
−t

n
X
Close
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X
ai āj
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1≤i6=j≤n
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=
=
n
X
i=1
n
X
X
|ai |2 +
1≤i<j≤n
|ai |2 + 2
i=1
X
ai āj +
ai āj
1≤j<i≤n
X
Re (ai āj )
1≤i<j≤n
and, similarly,
2
n
n
X
X
X
bk =
|bi |2 + 2
Re bi b̄j .
(2.113)
i=1
k=1
Also
n
X
ak
k=1
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
1≤i<j≤n
b̄k =
n
X
S.S. Dragomir
ai b̄i +
i=1
k=1
X
ai b̄j
Title Page
1≤i6=j≤n
and thus
(2.114)
Contents
Re
n
X
k=1
ak
n
X
!
b̄k
=
n
X
i=1
k=1
X
Re ai b̄i +
Re ai b̄j .
1≤i6=j≤n
Utilising (2.112) – (2.114), by (2.111), we deduce
" n
#
X
X
2
(2.115) f (t) =
|ak | + 2x
Re (ai āj ) t2
+2
II
I
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1≤i<j≤n
k=1
"
JJ
J
n
X
k=1
Page 51 of 288
#
Re ak b̄k + x
X
1≤i6=j≤n
Re ai b̄j
t
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+
n
X
X
|bk |2 + 2x
Re bi b̄j .
1≤i<j≤n
k=1
Since, by (2.110), f (t) ≥ 0 for any t ∈ R, it follows that the discriminant of
the quadratic function given by (2.115) is negative, i.e.,
1
0≥ ∆
4
" n
#2
X
X
Re ai b̄j
=
Re ak b̄k + x
1≤i6=j≤n
k=1
"
−
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
#"
2
|ak | + 2x
X
Re (ai āj )
1≤i<j≤n
k=1
n
X
#
2
|bk | + 2x
1≤i<j≤n
k=1
and the inequality (2.109) is proved.
(2.116)
n
X
k=1
#2
Re ak b̄k
Re bi b̄j
S.S. Dragomir
Title Page
Contents
Remark 2.19. If x = 0, then we get the (CBS) −inequality
"
X
≤
n
X
k=1
|ak |2
n
X
|bk |2 .
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References
[1] V.Y. BUNIAKOWSKI, Sur quelques inégalités concernant les intégrales
aux differences finies, Mem. Acad. St. Petersburg, (7) 1 (1859), No. 9,
1-18.
[2] A.L. CAUCHY, Cours d’Analyse de l’École Royale Polytechnique, I re Partie, Analyse Algébrique, Paris, 1821.
[3] P. SCHREIDER, The Cauchy-Bunyakovsky-Schwarz inequality, Hermann Graßmann (Lieschow, 1994), 64–70.
[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and
New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
[5] S.S. DRAGOMIR, On some inequalities (Romanian), “Caiete Metodico
Ştiinţifice”, No. 13, 1984, pp. 20. Faculty of Mathematics, Timişoara University, Romania.
[6] E.F. BECKENBACH AND R. BELLMAN, Inequalities, Springer-Verlag,
Berlin-Göttingen-Heidelberg, 1961.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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[7] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for
Real Numbers (Romanian), “Caiete Metodico-Ştiinţifice”, No. 57, pp. 24,
1989. Faculty of Mathematics, Timişoara University, Romania.
[8] S.S. DRAGOMIR AND J. SÁNDOR, Some generalisations of CauchyBuniakowski-Schwartz’s inequality (Romanian), Gaz. Mat. Metod.
(Bucharest), 11 (1990), 104–109.
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[9] I. CUCUREZEANU, Problems on Number Theory (Romanian), Ed. Technicǎ, Bucharest, 1976.
[10] S.S. DRAGOMIR, On an inequality of Tiberiu Popoviciu (Romanian),
Gaz. Mat. Metod., (Bucharest) 8 (1987), 124–128. ZBL No. 712:110A.
[11] T. POPOVICIU, Gazeta Matematicǎ, 16 (1940), p. 334.
[12] S.S. DRAGOMIR, Inequalities of Cauchy-Buniakowski-Schwartz’s type
for positive linear functionals (Romanian), Gaz. Mat. Metod. (Bucharest),
9 (1988), 162–164.
[13] T. ANDRESCU, D. ANDRICA AND M.O. DRÎMBE, The trinomial principle in obtaining inequalities (Romanian), Gaz. Mat. (Bucharest), 90
(1985), 332–338.
[14] P. FLOR, Über eine Unglichung von S.S. Wagner, Elemente Math., 20
(1965), 136.
[15] S.S. WAGNER, Amer. Math. Soc., Notices, 12 (1965), 220.
[16] S.S. DRAGOMIR, A version of Wagner’s inequality for complex numbers,
submitted.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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3.
3.1.
Refinements of the (CBS) −Inequality
A Refinement in Terms of Moduli
The following result was proved in [1].
Theorem 3.1. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of real
numbers. Then one has the inequality
(3.1)
n
X
k=1
a2k
n
X
k=1
b2k −
n
X
!2
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
ak b k
k=1
n
n
n
n
X
X
X
X
≥
ak |ak |
bk |bk | −
ak |bk |
|ak | bk ≥ 0.
k=1
k=1
k=1
k=1
Proof. We will follow the proof from [1].
For any i, j ∈ {1, . . . , n} the next elementary inequality is true:
(3.2)
|ai bj − aj bi | ≥ ||ai bj | − |aj bi || .
By multiplying this inequality with |ai bj − aj bi | ≥ 0 we get
(3.3)
(ai bj − aj bi )2
≥ |(ai bj − aj bi ) (|ai | |bj | − |aj | |bi |)|
= |ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi || .
S.S. Dragomir
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Summing (3.3) over i and j from 1 to n, we deduce
n
X
(ai bj − aj bi )2
i,j=1
n X
≥
a
|a
|
b
|b
|
+
b
|b
|
a
|a
|
−
|a
|
b
a
|b
|
−
a
b
|a
|
|b
|
i
i
j
j
i
i
j
j
i
i
j
j
i
j
j
i
i,j=1
n
X
≥
(ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi |) ,
i,j=1
giving the desired inequality (3.1).
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
The following corollary is a natural consequence of (3.1) [1, Corollary 4].
Corollary 3.2. Let ā be a sequence of real numbers. Then
!2 n
n
n
n
n
X
X
X
X
X
1
1
1
1
1
2
ak −
ak
≥
ak |ak | −
ak ·
|ak | ≥ 0.
(3.4)
n
n k=1
n k=1
n k=1
n k=1
k=1
There are some particular inequalities that may also be deduced from the
above Theorem 3.1 (see [1, p. 80]).
1. Suppose that for ā and b̄ sequences of real numbers, one has sgn (ak ) =
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sgn (bk ) = ek ∈ {−1, 1} . Then one has the inequality
(3.5)
n
X
a2k
k=1
n
X
b2k −
n
X
!2
ak b k
k=1
k=1
!2 ek ak bk ≥ 0.
k=1
n
n
X
X
≥ ek a2k
ek b2k −
k=1
k=1
n
X
A Survey on
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Type Discrete Inequalities
2. If ā = (a1 , . . . , a2n ) , then we have the inequality
(3.6)
2n
2n
X
"
a2k −
k=1
2n
X
#2
(−1)k ak
k=1
2n
2n
X
X
(−1)k |ak | ≥ 0.
≥
ak
k=1
k=1
3. If ā = (a1 , . . . , a2n+1 ) , then we have the inequality
(3.7)
(2n + 1)
2n+1
X
k=1
a2k −
2n+1
X
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J
!2
(−1)k ak
k=1
2n+1 2n+1
X
X
≥
ak
(−1)k |ak | ≥ 0.
k=1
S.S. Dragomir
k=1
The following version for complex numbers is valid as well.
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Theorem 3.3. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of
complex numbers. Then one has the inequality
(3.8)
n
X
2
|ai |
i=1
n
X
i=1
2
n
X
|bi | − ai b i i=1
n
n
n
n
X
X
X
X
≥
|ai | āi
|bi | bi −
|ai | bi
|bi | āi ≥ 0.
2
i=1
i=1
i=1
i=1
Proof. We have for any i, j ∈ {1, . . . , n} that
|āi bj − āj bi | ≥ ||ai | |bj | − |aj | |bi || .
Multiplying by |āi bj − āj bi | ≥ 0, we get
2
|āi bj − āj bi | ≥ ||ai | āi |bj | bj + |aj | āj |bi | bi − |ai | bi |bj | āj − |bi | āi |aj | bj | .
Summing over i and j from 1 to n and using the Lagrange’s identity for complex
numbers:
2
n
n
n
n
X
X
X
1X
2
2
|ai |
|bi | − ai bi =
|āi bj − āj bi |2
2 i,j=1
i=1
i=1
i=1
we deduce the desired inequality (3.8).
Remark 3.1. Similar particular inequalities may be stated, but we omit the
details.
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Type Discrete Inequalities
S.S. Dragomir
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3.2.
A Refinement for a Sequence Whose Norm is One
The following result holds [1, Theorem 6].
Theorem 3.4. Let ā = (a1 , . . . , an ), b̄ = P
(b1 , . . . , bn ) be sequences of real
numbers and ē = (e1 , . . . , en ) be such that ni=1 e2i = 1. Then the following
inequality holds
(3.9)
n
X
a2i
i=1
n
X
b2i
A Survey on
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i=1
n
#2
" n
n
n
n
X
X
X
X
X
≥ ak b k −
ek ak
ek bk + ek ak
ek bk k=1
k=1
k=1
k=1
k=1
!
2
n
X
≥
ak b k .
S.S. Dragomir
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k=1
Contents
Proof. We will follow the proof from [1].
From the (CBS) −inequality, one has
(3.10)
n
X
"
ak −
n
X
!
ei ai ek
i=1
k=1
≥
( n "
X
k=1
#2
n
X
JJ
J
"
bk −
ak −
i=1
#2
!
Go Back
ei bi ek
Close
i=1
k=1
n
X
n
X
!
ei ai ek
#"
bk −
n
X
II
I
!
ei bi ek
#)2
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.
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Since
Pn
2
k=1 ek
= 1, a simple calculation shows that
"
! #2
n
n
n
X
X
X
a2k −
ak −
ei ai ek =
i=1
k=1
n
X
"
n
X
bk −
#2
!
ei bi ek
=
i=1
k=1
n
X
k=1
k=1
n
X
n
X
b2k −
k=1
!2
,
ek ak
!2
ek bk
,
k=1
and
n
X
"
ak −
n
X
!
ei ai ek
#"
n
X
bk −
i=1
k=1
!
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Type Discrete Inequalities
#
ei bi ek
S.S. Dragomir
i=1
=
n
X
ak b k −
k=1
and then the inequality (3.10) becomes

!2   n
n
n
X
X
X
ek ak  
b2k −
(3.11) 
a2k −
k=1
k=1
k=1
≥
n
X
n
X
ek ak
k=1
n
X
ek bk
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k=1
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n
X
JJ
J
!2 
ek bk

Go Back
k=1
ak b k −
k=1
Using the elementary inequality
m2 − l2 p2 − q 2 ≤ (mp − lq)2 ,
II
I
n
X
k=1
ek ak
n
X
!2
ek bk
k=1
Close
≥ 0.
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m, l, p, q ∈ R
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for the choices
n
X
m=
! 12
a2k
,
k=1
n
X
and q = ek bk n
X
l=
ek ak ,
p=
k=1
n
X
! 12
b2k
k=1
k=1
the above inequality (3.11) provides the following result

(3.12)
n
X

! 12
a2k
k=1
n
X
! 12
b2k
k=1
2
n
n
X
X
−
ek ak
ek bk 
k=1
k=1
2
n
n
n
X
X
X
ek ak
ek bk .
≥
ak b k −
k=1
Since
n
X
k=1
! 21
a2k
n
X
k=1
! 21
b2k
k=1
S.S. Dragomir
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k=1
n
n
X
X
≥
ek ak
ek bk k=1
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Type Discrete Inequalities
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k=1
then, by taking the square root in (3.12) we deduce the first part of (3.9).
The second part is obvious, and the theorem is proved.
The following corollary is a natural consequence of the above theorem [1,
Corollary 7].
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P
Corollary 3.5. Let ā, b̄, ē be as in Theorem 3.4. If nk=1 ak bk = 0, then one
has the inequality:
!2
!2
n
n
n
n
X
X
X
X
b2k ≥ 4
ek ak
ek bk .
(3.13)
a2k
k=1
k=1
k=1
k=1
The following inequalities are interesting as well [1, p. 81].
1. For any ā, b̄ one has the inequality
(3.14)
n
X
k=1
a2k
n
X
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Type Discrete Inequalities
b2k
S.S. Dragomir
k=1
#2
" n
n
n
n
n
1 X
X
X
X
X
1
ak b k −
ak
bk + ak
bk ≥ n
n
k=1
k=1
k=1
k=1
k=1
!2
n
X
≥
ak b k .
k=1
2. If
Pn
k=1
(3.15)
ak bk = 0, then
n
X
k=1
a2k
n
X
k=1
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b2k
4
≥ 2
n
n
X
k=1
!2
ak
n
X
Close
!2
bk
.
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k=1
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In a similar manner, we may state and prove the following result for complex
numbers.
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Theorem 3.6. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be sequences of complex
numbers and
P ē = (e1 , . . . , en ) a sequence of complex numbers satisfying the
condition ni=1 |ei |2 = 1. Then the following refinement of the (CBS) −inequality
holds
(3.16)
n
X
i=1
2
|ai |
n
X
|bi |2
i=1
n
#2
" n
n
n
n
X
X
X
X
X
≥ ak b̄k −
ak ēk ·
ek b̄k + ak ēk ·
ek b̄k k=1
k=1
k=1
k=1
k=1
2
n
X
≥
ak b̄k .
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
k=1
The proof is similar to the one in Theorem 3.4 on using the corresponding
(CBS) −inequality for complex numbers.
Remark 3.2. Similar particular inequalities may be stated, but we omit the
details.
3.3.
A Second Refinement in Terms of Moduli
The following lemma holds.
Lemma 3.7. Let ā = (a1 , . . . , an ) be a sequence of P
real numbers and p̄ =
(p1 , . . . , pn ) a sequence of positive real numbers with ni=1 pi = 1. Then one
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has the inequality:
(3.17)
n
X
pi a2i −
n
X
i=1
!2
p i ai
i=1
n
n
n
X
X
X
≥
pi |ai | ai −
pi |ai |
p i ai .
i=1
i=1
i=1
Proof. By the properties of moduli we have
(ai − aj )2 = |(ai − aj ) (ai − aj )| ≥ |(|ai | − |aj |) (ai − aj )|
A Survey on
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for any i, j ∈ {1, . . . , n} . This is equivalent to
a2i − 2ai aj + a2j ≥ ||ai | ai + |aj | aj − |ai | aj − |aj | ai |
S.S. Dragomir
for any i, j ∈ {1, . . . , n} .
If we multiply (3.18) by pi pj ≥ 0 and sum over i and j from 1 to n we
deduce
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(3.18)
n
X
j=1
pj
n
X
pi a2i
−2
i=1
n
X
p i ai
i=1
≥
n
X
n
X
j=1
p j aj +
n
X
i=1
pi
n
X
pj a2j
j=1
pi pj |ai | ai + |aj | aj − |ai | aj − |aj | ai i,j=1
n
X
≥
pi pj (|ai | ai + |aj | aj − |ai | aj − |aj | ai ) ,
i,j=1
which is clearly equivalent to (3.17).
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Using the above lemma, we may prove the following refinement of the
(CBS) -inequality.
Theorem 3.8. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
real numbers. Then one has the inequality
!2
n
n
n
X
X
X
a2i
b2i −
ai b i
(3.19)
i=1
i=1
i=1
n
n
n
n
X
X
X
X
≥
a2i ·
sgn (ai ) |bi | bi −
|ai bi |
ai bi ≥ 0.
i=1
i=1
i=1
i=1
Proof. If we choose (for ai 6= 0, i ∈ {1, . . . , n}) in (3.17), that
a2
pi := Pn i
2
k=1 ak
, xi =
bi
, i ∈ {1, . . . , n} ,
ai
i=1
2
bi
Pn
·
−
2
ai
k=1 ak
n
X
a2
Pn i 2
≥
k=1 ak
i=1
a2i
n
X
i=1
a2i
bi
Pn
·
2
k=1 ak ai
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J
!2
n
bi bi X
a2i
· −
P
n
2
ai ai
k=1 ak
i=1
n
2
bi X
ai
bi P
· n
2
ai k=1 ak ai i=1
from where we get
n
X
S.S. Dragomir
Contents
we get
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
b2
Pn i
k=1
i=1
Pn |a |
Pn
Pn
Pn
i
2
|b
|
b
i i
( i=1 ai bi )
i=1
ai
i=1 |ai bi |
i=1 ai bi P
− Pn
−
P
n
2 ≥ 2
2
a2k
( k=1 a2k )
( nk=1 a2k )
k=1 ak
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which is clearly equivalent to (3.19).
The case for complex numbers is as follows.
Lemma 3.9. Let z̄ = (z1 , . . . , zn ) be a sequence of complex
P numbers and p̄ =
(p1 , . . . , pn ) a sequence of positive real numbers with ni=1 pi = 1. Then one
has the inequality:
2 n
n
n
n
n
X
X
X
X
X
(3.20)
pi |zi |2 − pi zi ≥ pi |zi | zi −
pi |zi |
pi zi .
i=1
i=1
i=1
i=1
i=1
Proof. By the properties of moduli for complex numbers we have
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
2
|zi − zj | ≥ |(|zi | − |zj |) (zi − zj )|
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for any i, j ∈ {1, . . . , n} , which is clearly equivalent to
2
2
|zi | − 2 Re (zi z̄j ) + |zj | ≥ ||zi | zi + |zj | zj − zi |zj | − |zi | zj |
for any i, j ∈ {1, . . . , n} .
If we multiply with pi pj ≥ 0 and sum over i and j from 1 to n, we deduce
the desired inequality (3.20).
Now, in a similar manner to the one in Theorem 3.8, we may state the following result for complex numbers.
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Theorem 3.10. Let ā = (a1 , . . . , an ) (ai 6= 0, i = 1, . . . , n) and b̄ = (b1 , . . . , bn )
be two sequences of complex numbers. Then one has the inequality:
2
n
n
n
X
X
X
(3.21)
|ai |2
|bi |2 − āi bi i=1
i=1
i=1
n
n
n
X
X
X
|a
|
i
|bi | bi −
|ai | bi
āi bi ≥ 0.
≥
ai
i=1
i=1
3.4.
i=1
A Refinement for a Sequence Less than the Weights
The following result was obtained in [1, Theorem 9] (see also [2, Theorem
3.10]).
Theorem 3.11. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be sequences of real
numbers and p̄ = (p1 , . . . , pn ), q̄ = (q1 , . . . , qn ) be sequences of nonnegative
real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then we have the
inequality
n
n
X
X
(3.22)
pk a2k
pk b2k
k=1
k=1
S.S. Dragomir
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
! 12 2
n
n
n
X
X
X
≥ 
(pk − qk ) ak bk +
qk a2k
qk b2k 
k=1
k=1
k=1
#2
!2
" n
n
n
X
X
X
p k ak b k .
≥ (pk − qk ) ak bk + q k ak b k ≥
k=1
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k=1
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Proof. We shall follow the proof in [1].
Since pk −qk ≥ 0, then the (CBS) −inequality for the weights rk := pk −qk
will give
(3.23)
n
X
pk a2k −
k=1
n
X
!
n
X
qk a2k
k=1
pk b2k −
n
X
k=1
!
qk b2k
k=1
"
≥
n
X
#2
(pk − qk ) ak bk
k=1
Using the elementary inequality
2
2
(ac − bd) ≥ a − b
.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
2
2
2
c −d
,
a, b, c, d ∈ R
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for the choices
a=
Contents
n
X
! 12
pk a2k
, b=
k=1
n
X
! 12
qk a2k
k=1
k=1
and
d=
n
X
k=1
, c=
n
X
! 21
qk b2k
! 12
pk b2k
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we deduce by (3.23) that

(3.24)
! 12
n
X
pk a2k

k=1
! 12
n
X
pk b2k
−
k=1
n
X
! 12
n
X
qk a2k
k=1
! 12 2
qk b2k

k=1
"
≥
n
X
#2
(pk − qk ) ak bk
.
k=1
Since, obviously,
n
X
! 21
pk a2k
k=1
n
X
! 12
pk b2k
≥
k=1
n
X
! 12
qk a2k
k=1
n
X
! 12
qk b2k
k=1
! 12
pk a2k
n
X
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! 12
JJ
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pk b2k
k=1
≥
n
X
k=1
S.S. Dragomir
k=1
then, by (3.24), on taking the square root, we would get
n
X
A Survey on
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Type Discrete Inequalities
! 12
qk a2k
n
X
k=1
! 12
qk b2k
n
X
+
(pk − qk ) ak bk ,
k=1
which provides the first inequality in (3.22).
The other inequalities are obvious and we omit the details.
The following corollary is a natural consequence of the above theorem [2,
Corollary 3.11].
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Corollary 3.12. Let ā, b̄ be sequences of real numbers and s̄ = (s1 , . . . , sn ) be
such that 0 ≤ sk ≤ 1 for any k ∈ {1, . . . , n} . Then one has the inequalities
(3.25)

! 21 2
n
n
n
X
X
X
sk a2k
sk b2k 
a2k
b2k ≥ 
(1 − sk ) ak bk +
k=1
k=1
k=1
k=1
k=1
#2
" n
n
X
X
≥ (1 − sk ) ak bk + s k ak b k k=1
k=1
!
2
n
X
≥
ak b k .
n
X
n
X
S.S. Dragomir
k=1
Remark 3.3. Assume that ā, b̄ and s̄ are as in Corollary 3.12. The following
inequalities hold (see [2, p. 15]).
P
a) If nk=1 ak bk = 0, then
n
X
(3.26)
a2k
k=1
b) If
Pn
k=1
(3.27)
A Survey on
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Type Discrete Inequalities
n
X
b2k ≥ 4
k=1
n
X
!2
s k ak b k
.
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k=1
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sk ak bk = 0, then
Quit

n
n
n
X
X
X
2
2

ak
bk ≥ ak b k +
k=1
k=1
k=1
n
X
k=1
αk a2k
n
X
k=1
! 21 2
αk b2k
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 .
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In particular, we may obtain the following particular inequalities involving
trigonometric functions (see [2, p. 15])
(3.28)
n
X
k=1
a2k
n
X
b2k
k=1

! 12 2
n
n
n
X
X
X
≥ 
ak bk cos2 αk +
a2k sin 2 αk
b2k sin2 αk 
k=1
k=1
k=1
#2
" n
n
X
X
2
2
≥ ak bk cos αk + ak bk sin αk k=1
k=1
!
2
n
X
≥
ak b k ,
k=1
(3.29)
k=1
a2k
n
X
b2k ≥ 4
k=1
!2
ak bk sin2 αk
.
k=1
k=1
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k=1
k=1
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Pn
ak bk sin2 αk = 0, then

n
n
n
X
X
X
2
2

(3.30)
ak
bk ≥ ak b k +
If
n
X
S.S. Dragomir
Contents
where ak , bk , αk ∈ R, k = 1, . . . , n.
P
If one would assume that nk=1 ak bk = 0, then
n
X
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n
X
k=1
a2k
2
sin αk
n
X
k=1
! 21 2
b2k sin2 αk
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 .
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3.5.
A Conditional Inequality Providing a Refinement
The following lemma holds [2, Lemma 4.1].
Lemma 3.13. Consider the sequences of real numbers x̄ = (x1 , . . . , xn ) , ȳ =
(y1 , . . . , yn ) and z̄ = (z1 , . . . , zn ) . If
(3.31)
yk2 ≤ |xk zk | for any k ∈ {1, . . . , n} ,
then one has the inequality:
n
X
(3.32)
!2
|yk |
≤
k=1
n
X
|xk |
n
X
A Survey on
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Type Discrete Inequalities
|zk | .
S.S. Dragomir
k=1
k=1
Proof. We will follow the proof in [2]. Using the condition (3.31) and the
(CBS) −inequality, we have
n
X
|yk | ≤
k=1
n
X
1
1
|xk | 2 |zk | 2
k=1
"
≤
n X
|xk |
1
2
n 2 X
k=1
=
n
X
|zk |
k=1
|xk |
k=1
which is clearly equivalent to (3.32).
n
X
k=1
! 21
|zk |
1
2
2
# 12
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The following result holds [2, Theorem 4.6].
Theorem 3.14. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) and c̄ = (c1 , . . . , cn ) be
sequences of real numbers such that
(i) |bk | + |ck | =
6 0 (k ∈ {1, . . . , n});
(ii) |ak | ≤
2|bk ck |
|bk |+|ck |
for any k ∈ {1, . . . , n} .
Then one has the inequality
(3.33)
P
P
n
X
2 nk=1 |bk | nk=1 |ck |
|ak | ≤ Pn
.
k=1 (|bk | + |ck |)
k=1
Proof. We will follow the proof in [2]. By (ii) we observe that

 2 |bk |
2 |bk ck |
for any k ∈ {1, . . . , n}
|ak | ≤
≤
|bk | + |ck | 
2 |ck |
and thus
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(3.34)
xk := 2 |bk | − |ak | ≥ 0
zk := 2 |ck | − |ak | ≥ 0
and
for any k ∈ {1, . . . , n} .
A simple calculation also shows that the relation (ii) is equivalent to
(3.35)
a2k ≤ (2 |bk | − |ak |) (2 |ck | − |ak |) for any k ∈ {1, . . . , n} .
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If we consider yk := ak and take xk , zk (k = 1, . . . , n) as defined by (3.34),
then we get yk2 ≤ xk zk (with xk , zk ≥ 0) for any k ∈ {1, . . . , n}. Applying
Lemma 3.13 we deduce
!2
!
!
n
n
n
n
n
X
X
X
X
X
(3.36)
|ak | ≤ 2
|bk | −
|ak |
2
|ck | −
|ak |
k=1
k=1
k=1
k=1
k=1
which is clearly equivalent to (3.33).
The following corollary is a natural consequence of the above theorem [2,
Corollary 4.7].
Corollary 3.15. For any sequence x̄ and ȳ of real numbers, with |xk |+|yk | =
6 0
(k = 1, . . . , n) , one has:
(3.37)
n
X
k=1
P
P
2 nk=1 |xk | nk=1 |yk |
|xk yk |
≤ Pn
.
|xk | + |yk |
k=1 (|xk | + |yk |)
For two positive real numbers, let us recall the following means
a+b
2
√
G (a, b) := ab
A (a, b) :=
and
H (a, b) :=
1
a
2
+
(the arithmetic mean)
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(the geometric mean)
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1
b
(the harmonic mean).
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We remark that if ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) are sequences of real
numbers, then obviously
!
n
n
n
X
X
X
(3.38)
A (ai , bi ) = A
ai ,
bi ,
i=1
i=1
i=1
n
X
n
X
and, by the (CBS) −inequality,
(3.39)
n
X
i=1
G (ai , bi ) ≤ G
i=1
ai ,
!
bi .
i=1
The following similar result for harmonic means also holds [2, p. 19].
Theorem 3.16. For any two sequences of positive real numbers ā and b̄ we
have the property:
!
n
n
n
X
X
X
(3.40)
H (ai , bi ) ≤ H
ai ,
bi .
i=1
i=1
i=1
Proof. Follows by Corollary 3.15 on choosing xk = ak , yk = bk and multiplying the inequality (3.37) with 2.
The following refinement of the (CBS) −inequality holds [2, Corollary 4.9].
This result is known in the literature as Milne’s inequality [8].
Theorem 3.17. For any two sequences of real numbers p̄ = (p1 , . . . , pn ), q̄ =
(q1 , . . . , qn ) with |pk | + |qk | =
6 0 (k = 1, . . . , n) , one has the inequality:
!
2
n
n
n
n
n
X
X
X
X
X
p2k qk2
2
(3.41)
p k qk
≤
p2k + qk2
≤
p
qk2 .
k
2
2
p
+
q
k
k=1
k=1
k=1 k
k=1
k=1
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Proof. We shall follow the proof in [2]. The first inequality is obvious by
p2 q 2
Lemma 3.13 on choosing yk = pk qk , xk = p2k +qk2 and zk = p2k+qk2 (k = 1, . . . , n) .
k
k
The second inequality follows by Corollary 3.15 on choosing xk = p2k and
y = qk2 (k = 1, . . . , n).
Remark 3.4. The following particular inequality is obvious by (3.41)
!2
n
n
X
X
(3.42)
sin αk cos αk
≤n
sin2 αk cos2 αk
i=1
≤
i=1
n
X
n
X
i=1
i=1
sin2 αk
cos2 αk ;
for any αk ∈ R, k ∈ {1, . . . , n} .
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3.6.
A Refinement for Non-Constant Sequences
The following result was proved in [3, Theorem 1].
Theorem 3.18. Let ā = (ai )i∈N , b̄ = (bi )i∈N , p̄ = (pi )i∈N be sequences of real
numbers such that
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(i) ai 6= aj and bi 6= bj for i 6= j, i, j ∈ N;
(ii) pi > 0 for all i ∈ N.
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Then for any H a finite part of N one has the inequality:
!2
X
X
X
(3.43)
pi a2i
pi b2i −
p i ai b i
≥ max {A, B} ≥ 0,
i∈H
i∈H
i∈H
where
i2
P
2
p
a
p
b
i∈H i i
j∈J j j
P
P
2
PJ i∈H pi a2i −
i∈J pi ai
hP
(3.44)
A := max
i∈H pi ai bi
J⊆H
J6=∅
P
j∈J pj aj −
P
and
hP
(3.45)
and PJ :=
B := max
i∈H
J⊆H
J6=∅
P
j∈J
p i ai b i
P
j∈J
p j bj −
PJ
P
i∈H
pi b2i −
2
i∈H pi bi
P
P
i∈J
p i bi
P
j∈J
p j aj
i2
2
S.S. Dragomir
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pj .
JJ
J
Proof. We shall follow the proof in [3].
Let J be a part of H. Define the mapping fJ : R → R given by


X
X
X
fJ (t) =
pi a2i 
pi b2i +
pi (bi + t)2 
i∈H
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i∈H\J
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2

−
X
i∈H\J
p i ai b i +
X
i∈J
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pi ai (bi + t) .
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Then by the (CBS) −inequality we have that fJ (t) ≥ 0 for all t ∈ R.
On the other hand we have
"
# "
#2
X
X
X
X
X
fJ (t) =
pi a2i
pi b2i + 2t
pi bi + t2 PJ −
p i ai b i + t
pi ai
i∈H
i∈H
i∈H
i∈J
i∈H

!2 
= t2 PJ
X
pi a2i −
i∈H
X
p i ai

i∈J
"
+ 2t
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#
X
pi a2i
i∈H
X
p i bi −
i∈J
p i ai b i
i∈H

X
X
+
pi a2i
pi b2i −
i∈H
X
i∈H
X
p i ai
i∈J
S.S. Dragomir
!2 
X
p i ai b i

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i∈H
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for all t ∈ R.
Since
!2
PJ
X
i∈H
pi a2i −
X
i∈J
p i ai
!2
≥ PJ
X
i∈J
pi a2i −
X
p i ai
>0
i∈J
as ai 6= aj for all i, j ∈ {1, . . . , n} with i 6= j, then, by the inequality fJ (t) ≥ 0
for any t ∈ R we get that
"
#2
X
X
X
X
1
p i ai b i
p j aj −
0≥ ∆=
pi a2i
p j bj
4
i∈H
j∈J
i∈H
j∈J
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!2 

− PJ
X
X
pi a2i −
i∈H
p i ai

i∈J

X
X

pi a2i
pi b2i −
i∈H
i∈H
!2 
X
pi ai bi

i∈H
from where results the inequality
A Survey on
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!2
X
pi a2i
i∈H
X
i∈H
pi b2i −
X
p i ai b i
≥ A.
i∈H
S.S. Dragomir
The second part of the proof goes likewise for the mapping gJ : R → R
given by
gJ (t) = 
Contents


X
pi a2i +
i∈H\J
X
i∈J
pi (ai + t)2 
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X
JJ
J
pi b2i
i∈H
2

−
X
p i ai b i +
i∈H\J
and we omit the details.
X
i∈J
pi bi (ai + t)
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The following corollary also holds [3, Corollary 1].
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Corollary 3.19. With the assumptions of Theorem 3.18 and if
2
P
P
P
P
2
i∈H pi ai bi
i∈H pi ai −
i∈H pi ai
i∈H pi bi
(3.46)
C :=
,
2
P
P
PH i∈H pi a2i −
p
a
i
i
i∈H
P
i∈H
D :=
(3.47)
p i ai b i
P
i∈H
p i bi −
PH
P
i∈H
pi b2i −
P
i∈H
pi b2i
P
i∈H
P
p i bi
i∈H
p i ai
2
2
,
then one has the inequality
A Survey on
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!2
X
(3.48)
pi a2i
i∈H
X
pi b2i −
i∈H
X
≥ max {C, D} ≥ 0.
p i ai b i
The following corollary also holds [3, Corollary 2].
Corollary 3.20. If ai , bi 6= 0 for i ∈ N and H is a finite part of N, then one has
the inequality
!2
X
X
X
2
2
(3.49)
p i ai
p i bi −
p i ai b i
i∈H
i∈H
1
max
card (H) − 1
(P
)
P
2
2
j∈H pj cj
j∈H pj dj
P
,P
≥ 0,
2
2
i∈H pi ai
i∈H pi bi
where
(3.50)
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i∈H
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cj := aj
X
i∈H
p i ai b i − b j
X
i∈H
pi a2i ,
j∈H
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and
dj := aj
(3.51)
X
X
pi b2i − bj
i∈H
pi ai bi , j ∈ H.
i∈H
Proof. Choosing in Theorem 3.18, J = {j} , we get the inequality
!2
X
i∈H
pi a2i
X
pi b2i
−
i∈H
X
p i ai b i
i∈H
p2j c2j
≥ P
, j∈H
pj i∈H pi a2i − p2j a2j
from where we obtain
!
X
X
X
pi a2i − pj a2j 
pi a2i
pi b2i −
i∈H
i∈H
i∈H
p i ai b i
i∈H
 ≥ pj c2j for any j ∈ H.
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i∈H
Summing these inequalities over j ∈ H, we get

X
X
X
[card (H) − 1]
pi a2i 
pi a2i
pi b2i −
i∈H
S.S. Dragomir
!2 
X
i∈H
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!2 
X
p i ai b i
i∈H
≥
X
pj c2j
j∈H
from where we get the first part of (3.49).
The second part goes likewise and we omit the details.
Remark 3.5. The following particular inequalities provide refinement for the
(CBS) −inequality [3, p. 60 – p. 61].
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1. Assume that ā = (a1 , . . . , an ), b = (b1 , . . . , bn ) are nonconstant sequences of real numbers. Then
(3.52)
n
X
a2i
i=1
n
X
b2i −
n
X
!2
ai b i
i=1
i=1



 [Pn a2 Pn b − Pn a Pn a b ]2
i=1 i
i=1 i
i=1 i
i=1 i i
,
≥ max
n
P
P

2
n
2

n ai − ( i=1 ai )

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i=1
[
Pn
i=1 bi
Pn
n
i=1
n
P
i=1
Pn
Pn
ai bi − i=1 ai
P
2
b2i − ( ni=1 bi )
2
i=1 bi ]


2

S.S. Dragomir
.



2. Assume that ā and b̄ are sequences of real numbers with not all elements
equal to zero, then
(3.53)
n
X
i=1
a2i
n
X
b2i −
i=1
1
≥
max
n−1
n
X
!2
ai b i
(i=1
Pn
j=1
Pn
Pn 2 2
i=1 ai bi − bj
i=1 ai )
Pn 2
,
i=1 ai
)
Pn
Pn 2
Pn
2
(a
a
−
b
a
b
)
j
i
i
j
j=1
i=1 i
i=1
Pn 2
.
b
i=1 i
(aj
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3.7.
De Bruijn’s Inequality
The following refinement of the (CBS) −inequality was proved by N.G. de
Bruijn in 1960, [4] (see also [5, p. 89]).
Theorem 3.21. If ā = (a1 , . . . , an ) is a sequence of real numbers and z̄ =
(z1 , . . . , zn ) is a sequence of complex numbers, then
#
2
" n
n
n
n
X
X
1X 2 X
ak
|zk |2 + zk2 .
(3.54)
ak zk ≤
2
k=1
k=1
k=1
k=1
Equality holds in (3.54) if and only
Pifnfor k2 ∈2 {1, . . . , n} , ak = Re (λzk ) , where
λ is a complex number such that k=1 λ zk is a nonnegative real number.
Proof. We shall follow the proof in [5, p. 89 – p. 90].
By a simultaneous rotation of all the zk ’s about the origin, we get
n
X
S.S. Dragomir
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JJ
J
ak zk ≥ 0.
k=1
This rotation does not affect the moduli
n
n
X
X ak zk , zk2 and |zk | for k ∈ {1, . . . , n} .
k=1
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k=1
Hence, it is sufficient to prove inequality (3.54) for the case where
0.
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k=1
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ak zk ≥
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If we put zk = xk + iyk (k ∈ {1, . . . , n}) , then, by the (CBS) −inequality
for real numbers, we have
2
!2
n
n
n
n
X
X
X
X
2
x2k .
ak zk
≤
ak
(3.55)
ak zk =
k=1
k=1
k=1
k=1
Since
2x2k = |zk |2 + Re zk2 for any k ∈ {1, . . . , n}
we obtain, by (3.55), that
2
" n
#
n
n
n
X
X
X
X
1
(3.56)
ak zk ≤
a2k
|zk |2 +
Re zk2 .
2 k=1
k=1
k=1
k=1
As
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n
X
k=1
Re zk2 = Re
n
X
k=1
!
zk2
n
X
≤
zk2 ,
k=1
then by (3.56) we deduce the desired inequality (3.54).
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3.8.
McLaughlin’s Inequality
The following refinement of the (CBS) −inequality for sequences of real numbers was obtained in 1966 by H.W. McLaughlin [7, p. 66].
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Theorem 3.22. If ā = (a1 , . . . , a2n ), b̄ = (b1 , . . . , b2n ) are sequences of real
numbers, then
!2 " n
#2
2n
2n
2n
X
X
X
X
2
(3.57)
ai b i +
(ai bn+i − an+i bi ) ≤
ai
b2i
i=1
i=1
i=1
i=1
with equality if and only if for any i, j ∈ {1, . . . , n}
ai bj − aj bi − an+i bn+j + an+j bn+i = 0
(3.58)
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and
S.S. Dragomir
ai bn+j − aj bn+i + an+i bj − an+j bi = 0.
(3.59)
Proof. We shall follow the proof in [6] by M.O. Drîmbe.
The following identity may be obtained by direct computation
(3.60)
2n
X
i=1
a2i
2n
X
b2i
−
i=1
=
2n
X
!2
ai b i
"
−
i=1
X
n
X
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#2
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(ai bn+i − an+i bi )
i=1
(ai bj − aj bi − an+i bn+j + an+j bn+i )2
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+
X
2
(ai bn+j − aj bn+i + an+i bj − an+j bi ) .
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1≤i<j≤n
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It is obvious that (3.57) is a simple consequence of the identity (3.60). The case
of equality is also obvious.
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Remark 3.6. For other similar (CBS) −type inequalties see the survey paper
[7]. An analogous inequality to (3.57) for sequences ā and b̄ having 4n terms
each may be found in [7, p. 70].
3.9.
A Refinement due to Daykin-Eliezer-Carlitz
We will present now the version due to Mitrinović, Pečarić and Fink [5, p. 87]
of Daykin-Eliezer-Carlitz’s refinement of the discrete (CBS) −inequality [8].
Theorem 3.23. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
positive numbers. The ineuality
!2
n
n
n
n
n
X
X
X
X
X
2
(3.61)
ai b i
≤
f (ai bi )
g (ai bi ) ≤
ai
b2i
i=1
i=1
i=1
i=1
holds if and only if
f (a, b) g (a, b) = a b ,
f (ka, kb) = k 2 f (a, b) ,
bf (a, 1) af (b, 1)
a b
+
≤ +
af (b, 1) bf (a, 1)
b a
(3.64)
S.S. Dragomir
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2 2
(3.62)
(3.63)
i=1
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for any a, b, k > 0.
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Proof. We shall follow the proof in [5, p. 88 – p. 89].
Necessity. Indeed, for n = 1, the inequality (3.61) becomes
(ab)2 ≤ f (a, b) g (a, b) ≤ a2 b2 ,
a, b > 0
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which gives the condition (3.62).
For n = 2 in (3.61), using (3.62), we get
2a1 b1 a2 b2 ≤ f (a1 , b1 ) g (a2 , b2 ) + f (a2 , b2 ) g (a1 , b1 ) ≤ a21 b22 + a22 b21 .
By eliminating g, we get
(3.65)
2≤
f (a1 , b1 ) a2 b2 f (a2 , b2 ) a1 b1
a1 b 2 a2 b 1
·
+
·
≤
+
.
f (a2 , b2 ) a1 b1 f (a1 , b1 ) a2 b2
a2 b 1 a1 b 2
By substituting in (3.65) a, b for a1 , b1 and ka, kb for a2 , b2 (k > 0), we get
f (a, b) 2 f (ka, kb) −2
2≤
k +
k ≤2
f (ka, kb)
f (a, b)
and this is valid only if k 2 f (a, b) (f (ka, kb)) = 1, i.e., the condition (3.63)
holds.
Using (3.65), for a1 = a, b1 = b, a2 = b, b2 = 1, we have
2≤
(3.66)
f (a,1)
a
f (b,1)
b
+
f (b,1)
b
f (a,1)
a
≤
a b
+ .
b a
The first inequality in (3.66) is always satisfied while the second inequality is
equivalent to (3.64).
Sufficiency. Suppose that (3.62) holds. Then inequality (3.61) can be written in
the form
X
X
2
ai b i aj b j ≤
[f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi )]
1≤i<j≤n
1≤i<j≤n
≤
X
1≤i<j≤n
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a2i b2j + a2j b2i .
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Therefore, it is enough to prove
(3.67)
2ai bi aj bj ≤ f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi )
≤ a2i b2j + a2j b2i .
Suppose that (3.64) holds. Then (3.66) holds and putting a =
(3.66) and using (3.63), we get
2≤
ai
,
bi
b =
aj
bj
in
f (ai , bi ) aj bj f (aj , bj ) ai bi
ai bj aj bi
·
+
·
≤
+
.
f (aj , bj ) ai bi
f (ai , bi ) aj bj
aj b i ai b j
A Survey on
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Multiplying the last inequality by ai bi aj bj and using (3.62), we obtain (3.67).
S.S. Dragomir
Remark 3.7. In [8] (see [5, p. 89]) the condition (3.64) is given as
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(3.68)
f (b, 1) ≤ f (a, 1) ,
f (b, 1)
f (a, 1)
≤
for a ≥ b > 0.
2
a
b2
Remark 3.8. O.E. Daykin, C.J. Eliezer and C. Carlitz [8] stated that examples
for f, g satisfying (3.62) – (3.64) were obtained in the literature. The choice
2 y2
f (x, y) = x2 + y 2 , g (x, y) = xx2 +y
2 will give the Milne’s inequality
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(3.69)
n
X
i=1
!2
ai bi
≤
n
X
i=1
n
n
n
X
X
X
a2i b2i
2
2
2
ai + b i ·
≤
ai ·
b2i .
2
2
a
+
b
i
i=1 i
i=1
i=1
For a different proof of this fact, see Section 3.5.
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The choice f (x, y) = x1+α y 1−α , g (x, y) = x1−α y 1+α (α ∈ [0, 1]) will give
the Callebaut inequality
!2
n
n
n
n
n
X
X
X
X
X
1+α 1−α
1−α 1+α
2
(3.70)
ai b i
≤
ai b i
ai b i ≤
ai ·
b2i .
i=1
3.10.
i=1
i=1
i=1
i=1
A Refinement via Dunkl-Williams’ Inequality
We will use the following version of Dunkl-Williams’ inequality established in
1964 in inner product spaces [9].
Lemma 3.24. Let a, b be two non-null complex numbers. Then
a
1
b (3.71)
|a − b| ≥ (|a| + |b|) − .
2
|a| |b|
Proof. We start with the identity (see also [5, pp. 515 – 516])
2 a
b
ā
b̄
− = a − b
−
|a| |b| |a| |b|
|a| |b|
a b̄
= 2 − 2 Re
·
|a| |b|
1
=
2 |a| |b| − 2 Re a · b̄
|a| |b|
1 2 |a| |b| − |a|2 + |b|2 − |a − b|2
=
|a| |b|
1 =
|a − b|2 − (|a| − |b|)2 .
|a| |b|
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Hence
2
2 a
1
b
(|a| − |b|)2 |a − b| −
(|a| + |b|) − =
(|a| + |b|)2 − |a − b|2
2
|a| |b|
4 |a| |b|
2
and (3.71) is proved.
Using the above result, we may prove the following refinement of the (CBS) −
inequality for complex numbers.
Theorem 3.25. If ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) are two sequences of
nonzero complex numbers, then
2
n
n
n
X
X
X
(3.72)
|ak |2
|bk |2 − ak b k k=1
k=1
k=1
2
n 1 X |bi |
|aj |
|ai |
|bj | ≥
āi bj − āj bi +
āi ·
bj −
bi ·
āj ≥ 0.
8 i,j=1 |ai |
|bj |
|bi |
|aj | Proof. The inequality (3.71) is clearly equivalent to
2
1 |b|
|a| 2
·a−
·b
(3.73)
|a − b| ≥ a − b +
4
|a|
|b| for any a, b ∈ C, a, b 6= 0.
We know the Lagrange’s identity for sequences of complex numbers
2
n
n
n
n
X
X
X
1X
ak b k =
(3.74)
|ak |2
|bk |2 − |āi bj − āj bi |2 .
2
i,j=1
k=1
k=1
k=1
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By (3.73), we have
2
1 |aj | |bi |
|ai | |bj |
|āi bj − āj bi | ≥ āi bj − āj bi +
āi bj −
āj bi .
4
|ai | |bj |
|aj | |bi |
2
Summing over i, j from 1 to n and using the (CBS) −inequality for double
sums, we deduce (3.72).
3.11.
Some Refinements due to Alzer and Zheng
In 1992, H. Alzer [10] presented the following refinement of the Cauchy-Schwarz
inequality written in the form
!2
n
n
n
X
X
X
(3.75)
xk yk
≤
yk
x2k yk .
k=1
k=1
k=1
Theorem 3.26. Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 =
x0 < x1 ≤ x22 ≤ · · · ≤ xnn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then
(3.76)
n
X
k=1
!2
xk yk
n X
1
2
≤
yk
xk − xk−1 xk yk ,
4
k=1
k=1
n
X
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with equality holding if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn .
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In 1998, Liu Zheng [11] pointed out an error in the proof given in [10],
which can be corrected as shown in [11]. Moreover, Liu Zheng established the
following result which sharpens (3.76).
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Theorem 3.27. Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 <
x1 ≤ x22 ≤ · · · ≤ xnn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then
(3.77)
n
X
!2
≤
xk yk
k=1
n
X
k=1
yk
n
X
δk yk ,
k=1
with
(3.78)
δ1 = x21 and δk =
7k + 1 2
k
xk −
x2k−1
8k
8 (k − 1)
(k ≥ 2) .
Equality holds in (3.77) if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · =
yn .
In 1999, H. Alzer improved the above results as follows.
To present his results, we will follow [12].
In order to prove the main result, we need some technical lemmas.
Lemma 3.28. Let xk (k = 1, . . . , n) be real numbers such that
0 < x1 ≤
x2
xn
≤ ··· ≤
.
2
n
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Then
(3.79)
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2
n
X
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xk ≤ (n + 1) xn ,
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with equality holding if and only if xk = kx1 (k = 1, . . . , n) .
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A proof of Lemma 3.28 is given in [10].
Lemma 3.29. Let xk (k = 1, . . . , n) be real numbers such that
xn
x2
0 < x1 ≤
≤ ··· ≤
.
2
n
Then
!2
n
n
X
X
3k + 1 2
≤n
(3.80)
xk
xk ,
4k
k=1
k=1
A Survey on
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with equality holding if and only if xk = kx1 (k = 1, . . . , n) .
S.S. Dragomir
Proof. Let
Sn = Sn (x1 , . . . , xn ) = n
n
X
3k + 1
k=1
4k
x2k −
n
X
!2
xk
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k=1
Then we have for n ≥ 2 :
(3.81)
Sn − Sn−1 =
n−1
X
3k + 1
k=1
4k
x2k
− 2xn
n−1
X
k=1
xk +
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xn
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= f (xn ) , say.
We differentiate with respect to xn and use (3.79) and xn ≥
yields
n−1
X
3 (n − 1)
n
0
f (xn ) =
xn − 2
xk ≥ xn−1 > 0
2
2
k=1
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n
x .
n−1 n−1
This
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and
(3.82)
f (xn ) ≥ f
=
n
xn−1
n−1
n
X
3k + 1
k=1
4k
x2k
n−1
X
2n
3n2
xk +
−
xn−1
x2n−1
n−1
4 (n − 1)
k=1
= Tn−1 (x1 , . . . , xn−1 ) ,
say.
We use induction on n to establish that Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2.
We have T1 (x1 ) = 0. Let n ≥ 3; applying (3.79) we obtain
A Survey on
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n−1
3n + 2
∂
2n X
Tn−1 (x1 , . . . , xn−1 ) =
xn−1 −
xk
∂xn−1
2
n − 1 k=1
≥
(n − 2) (n + 1)
xn−1 > 0
2 (n − 1)
and
(3.83)
Tn−1 (x1 , . . . , xn−1 ) ≥ Tn−1
n−1
x1 , . . . , xn−2 ,
xn−2 .
n−2
Using the induction hypothesis Tn−2 (x1 , . . . , xn−2 ) ≥ 0 and (3.79), we get
n−1
(3.84)
Tn−1 x1 , . . . , xn−2 ,
xn−2
n−2
"
#
n−2
X
xn−2
≥
(n − 1) xn−2 − 2
xk .
n−2
k=1
S.S. Dragomir
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From (3.83) and (3.84) we conclude Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2, so that
(3.81) and (3.82) imply
(3.85)
Sn ≥ Sn−1 ≥ · · · ≥ S2 ≥ S1 = 0.
This proves inequality (3.80). We discuss the cases of equality. A simple calculation reveals that Sn (x1 , 2x1 , . . . , nx1 ) = 0. We use induction on n to prove
the implication
(3.86)
Sn (x1 , . . . , xn ) = 0 =⇒ xk = kx1 for k = 1, . . . , n.
If n = 1, then (3.86) is obviously true. Next, we assume that (3.86) holds
with n − 1 instead of n. Let n ≥ 2 and Sn (x1 , . . . , xn ) = 0. Then (3.85)
leads to Sn−1 (x1 , . . . , xn−1 ) = 0 which implies xk = kx1 for k = 1, . . . , n −
1. Thus, we have Sn (x1 , 2x1 , . . . , (n − 1) x1 , xn ) = 0 which is equivalent to
(xn − nx1 ) (3xn − nx1 ) = 0. Since 3xn > nx1 , we get xn = nx1 .
Lemma 3.30. Let xk (k = 1, . . . , n) be real numbers such that
0 < x1 ≤
xn
x2
≤ ··· ≤
.
2
n
If the natural numbers n and q satisfy n ≥ q + 1, then
!2
q
q
X
X
(3n + 1) q 2 2
xk − 2qxn
xk +
(3.87)
0<
xn .
4n
k=1
k=1
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Proof. We denote the expression on the right-hand side of (3.87)byu (xn ) .
Then we differentiate with respect to xn and apply (3.79), xn ≥ nq xq and
n ≥ q + 1. This yields
q
X
1 0
(3n + 1) q
u (xn ) =
xn − 2q
xk
q
2n
k=1
(3n + 1) q
xn − (q + 1) xq
2n
3n − 2q − 1
≥
xq > 0.
2
≥
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Hence, we get
(3.88)
u (xn ) ≥ u
n
xq
q
=
(3n + 1) n 2
xq − 2nxq
4
Let
q
X
xk +
k=1
q
X
!2
xk
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.
k=1
q
v (t) =
X
(3t + 1) t
xq − 2t
xk and t ≥ q + 1;
4
k=1
from (3.79) we conclude that
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q
v 0 (t) =
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X
6t + 1
2q + 3
xq − 2
xk ≥
xq > 0.
4
4
k=1
This implies that the expression on the right-hand side of (3.88) is increasing on
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[q + 1, ∞) with respect to n. Since n ≥ q + 1, we get from (3.88):
q
(3.89)
X
(3q + 4) (q + 1) 2
u (xn ) ≥
xq − 2 (q + 1) xq
xk +
4
k=1
q
X
!2
xk
k=1
= Pq (x1 , . . . , xq ) , say.
We use induction on q to show that Pq (x1 , . . . , xq ) > 0 for q ≥ 1. We have
P1 (x1 ) = 12 x21 . If Pq−1 (x1 , . . . , xq−1 ) > 0, then we obtain for q ≥ 2 :
(3.90) Pq (x1 , . . . , xq ) > 2q (xq−1 − xq )
q−1
X
k=1
xk −
(3q + 1) q 2
xq−1
4
q (3q − 1) 2
+
xq = w (xq ) , say.
4
q
xq−1 . Then
We differentiate with respect to xq and use (3.79) and xq ≥ q−1
we get
#
"
q−1
X
q 2 (q + 1)
3q
−
1
xq − 2
xk ≥
xq−1 > 0
w0 (xq ) = q
2
2
(q
−
1)
k=1
and
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q
w (xq ) ≥ w
xq−1
q−1
"
#
q−1
2
X
4q − q − 1
q
xq−1
xq−1 − 2
xk
=
q−1
4 (q − 1)
k=1
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≥
(3q − 1) q 2
x
> 0.
4 (q − 1) q−1
From (3.89), (3.90) and (3.91), we obtain u (xn ) > 0.
We are now in a position to prove the following companion of inequalities
(3.76) and (3.77) (see [12]).
Theorem 3.31. The inequality
!2
n n
n
X
X
X
β
xk yk
≤
yk
α+
(3.92)
yk ,
k
k=1
k=1
k=1
holds for all natural numbers n and for all real numbers xk and yk (k = 1, . . . , n)
with
(3.93)
0 < x1 ≤
xn
x2
≤ ··· ≤
and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 ,
2
n
if and only if
α≥
3
and β ≥ 1 − α.
4
Proof. First, we assume that (3.92) is valid for all n ≥ 1 and for all real numbers
xk and yk (k = 1, . . . , n) which satisfy (3.93). We set xk = k and yk = 1
(k = 1, . . . , n). Then (3.92) leads to
3
3
2n + α + 3β −
(n ≥ 1) .
(3.94)
0≤ α−
4
2
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This implies α ≥ 34 . And, (3.94) with n = 1 yields α + β ≥ 1.
Now, we suppose that α ≥ 34 and β ≥ 1 − α. Then we obtain for k ≥ 1 :
α+
β
1−α
3
1
≥α+
≥ + ,
k
k
4 4k
so that is suffices to show that (3.92) holds with α =
F (y1 , . . . , yn ) =
n
X
k=1
yk
n
X
3k + 1
k=1
4k
3
4
x2k yk −
and β = 14 . Let
n
X
!2
xk yk
k=1
and
S.S. Dragomir
Fq (y) = F (y, . . . , y, yq+1 , . . . , yn )
(1 ≤ q ≤ n − 1) .
We shall prove that Fq is strictly increasing on [yq+1 , ∞). Since yq+1 ≤ yq , we
obtain
(3.95)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
Fq (yq ) ≥ Fq (yq+1 ) = Fq+1 (yq+1 )
(1 ≤ q ≤ n − 1) ,
and Lemma 3.29 imply
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F (y1 , . . . , yn ) = F1 (y1 ) ≥ F1 (y2 ) = F2 (y2 ) ≥ F2 (y3 )
≥ · · · ≥ Fn−1 (yn−1 ) ≥ Fn−1 (y)

!2 
n
n
X
X
3k + 1 2
= yn2 n
xk −
xk  ≥ 0.
4k
k=1
k=1
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If F (y1 , . . . , yn ) = 0, then we conclude from the strict monotonicity of Fq and
from Lemma 3.29 that y1 = · · · = yn and xk = kx1 (k = 1, . . . , n) .
It remains to show that Fq is strictly increasing on [yq+1 , ∞). Let y ≥ yq+1 ;
we differentiate Fq and apply Lemma 3.29. This yields

!2 
q
q
n
X
X
X
3k + 1 2
3k + 1 2
0


Fq (y) = 2y q
xk
+q
xk −
xk yk
4k
4k
k=1
k=1
k=q+1
+
n
X
k=q+1
yk
q
X
3k + 1
k=1
4k
x2k − 2
n
X
k=q+1
xk yk
q
X
xk
k=1
S.S. Dragomir
and
q
X 3k + 1
1 00
Fq (y) = q
x2k −
2
4k
k=1
!2
q
X
xk
≥ 0.
k=1
Title Page
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Hence, we have
(3.96)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
Fq0 (y)≥ Fq0 (yq+1 )
! q
!2 
q
n
X
X 3k + 1
X
1
= 2qyq+1 +
yk 
x2k −
xk 
4k
q
k=q+1
k=1
k=1

!2 
q
q
n


2
X
X
1 X
(3k + 1) q 2
+
yk
xk −2qxk
xi +
xi
.

q k=q+1 
4k
i=1
i=1
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Fq0
(yq+1 ) > 0, so that (3.96)
From Lemma 3.29 and Lemma 3.30 we obtain
implies Fq0 (y) > 0 for y ≥ yq+1 . This completes the proof of the theorem.
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Remark 3.9. The proof of the theorem reveals that the sign of equality holds
in (3.92) (with α = 34 and β = 14 ) if and only if xk = kx1 (k = 1, . . . , n) and
y1 = · · · = yn .
Remark 3.10. If δk is defined by (3.78), then we have for k ≥ 2 :
"
2 #
x 2 x
3
1
k
(k
−
1)
k
k−1
δk −
+
x2k =
−
,
4 4k
8
k
k−1
which implies that inequality (3.92) (with α =
3
4
and β = 14 ) sharpens (3.77).
Remark 3.11. It is shown in [10] thatif a sequence (xk ) satisfies x0 = 0 and
2xk ≤ xk−1 + xk+1 (k ≥ 1) , then xkk is increasing. Hence, inequality (3.92)
is valid for all sequences (xk ) which are positive and convex.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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References
[1] S.S. DRAGOMIR AND N.M. IONESCU, Some refinements of CauchyBuniakowski-Schwartz inequality for sequences, Proc. of the Third Symp.
of Math. and Its Appl., 3-4 Nov. 1989, Polytechnical Institute of Timişoara,
Romania, 78–82.
[2] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s inequality for real
numbers (Romanian), “Caiete Metodico Ştiinţifice”, No. 57, 1989, pp. 24.
Faculty of Mathematics, Timişoara University, Romania.
[3] S.S. DRAGOMIR, Š.Z. ARSLANAGIĆ AND D.M. MILOŠEVIĆ, On
Cauchy-Buniakowski-Schwartz’s inequality for real numbers, Mat. Bilten,
18 (1994), 57–62.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
[4] N.G. de BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12–14.
Title Page
[5] D.S. MITRINOVIĆ, J.E. PEC̆ARIĆ AND A.M. FINK, Classical and
New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
Contents
[6] M.O. DRÎMBE, A generalisation of Lagrange’s identity (Romanian), Gaz.
Mat. (Bucharest), 89 (1984), 139-141.
[7] H.W. McLAUGHLIN, Inequalities complementary to the CauchySchwartz inequality for finite sums of quaternions, refinements of the
Cauchy-Schwartz inequality for finite sums of real numbers; inequalities
concerning ratios and differences of generalised means of different order,
Maryland University, Technical Note BN-454, (1966), 129 pp.
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[8] D.E. DAYKIN, C.J. ELIEZER AND C. CARLITZ, Problem 5563, Amer.
Math. Monthly, 75 (1968), p. 198 and 76 (1969), 98–100.
[9] C.F. DUNKL AND K.S. WILLIAMS, A simple inequality, Amer. Math.
Monthly, 71 (1964), 53–54.
[10] H. ALZER, A refinement of the Cauchy-Schwartz inequality, J. Math.
Anal. Appl., 168 (1992), 596–604.
[11] L. ZHENG, Remark on a refinement of the Cauchy-Schwartz inequality,
J. Math. Anal. Appl., 218 (1998), 13–21.
[12] H. ALZER, On the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 234
(1999), 6–14.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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4.
4.1.
Functional Properties
A Monotonicity Property
The following result was obtained in [1, Theorem].
Theorem 4.1. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be sequences of real
numbers and p̄ = (p1 , . . . , pn ) , q̄ = (q1 , . . . , qn ) be sequences of nonnegative
real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then one has the
inequality
(4.1)
n
X
! 21
pi a2i
n
X
i=1
i=1
! 12
n
X
pi b2i
−
p i ai b i i=1
! 12 n
! 12
n
n
X
X
X
qi b2i
−
qi ai bi ≥ 0.
≥
qi a2i
i=1
i=1
i=1
Proof. We shall follow the proof in [1].
Since pk − qk ≥ 0, then the (CBS) −inequality for the weights rk = pk − qk
(k ∈ {1, . . . , n}) will produce
! n
! " n
#2
n
n
n
X
X
X
X
X
(4.2)
pk a2k −
qk a2k
pk b2k −
qk b2k ≥
(pk −qk ) ak bk .
k=1
k=1
k=1
k=1
k=1
S.S. Dragomir
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Using the elementary inequality
(ac − bd)2 ≥ a2 − b2
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
c2 − d2 ,
a, b, c, d ∈ R
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for the choices
a=
n
X
! 12
pk a2k
,
b=
k=1
d=
n
X
n
X
! 21
qk a2k
,
c=
k=1
n
X
! 12
pk b2k
and
k=1
! 21
qk b2k
k=1
we deduce by (4.2), that
n
X
! 21
n
X
pk a2k
k=1
! 12
pk b2k
k=1
n
X
≥
−
p k ak b k −
k=1
n
X
! 12
qk a2k
k=1
n
X
k=1
n
X
k=1
n
X
q k ak b k ≥ k=1
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
! 12
qk b2k
S.S. Dragomir
n
X
p k ak b k − qk ak bk Title Page
k=1
proving the desired inequality (4.1).
The following corollary holds [1, Corollary 1].
Corollary 4.2. Let ā and b̄ be as in Theorem 4.1. Denote
Sn (1) := {x̄ = (x1 , . . . , xn ) |0 ≤ xi ≤ 1, i ∈ {1, . . . , n}} .
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Then
(4.3)
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n
X
i=1
! 12
a2i
n
X
i=1
! 12
b2i
n
X
−
ai b i i=1
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
= sup 
n
X
x̄∈Sn (1)
! 12
n
X
xi a2i
i=1
! 12
xi b2i
i=1

n
X
−
xi ai bi  ≥ 0.
i=1
Remark 4.1. The following inequality is a natural particular case that may be
obtained from (4.1) [1, p. 79]
n
X
(4.4)
! 21
a2i
i=1
"
≥
n
X
i=1
! 12
n
X
b2i
i=1
i=1
# 12 "
a2i
2
n
X
−
ai b i trig (αi )
n
X
# 12
b2i trig 2 (αi )
i=1
n
X
2
−
ai bi trig (αi ) ≥ 0,
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
i=1
where trig (x) = sin x or cos x, x ∈ R and ᾱ = (α1 , . . . , αn ) is a sequence of
real numbers.
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4.2.
A Superadditivity Property in Terms of Weights
Let Pf (N) be the family of finite parts of the set of natural numbers N, S (K)
the linear space of real or complex numbers, i.e.,
S (K) := x|x = (xi )i∈N , xi ∈ K, i ∈ N
and S+ (R) the family of nonnegative real sequences. Define the mapping
! 12 X
X
X
(4.5)
S (p, I, x, y) :=
pi |xi |2
pi |yi |2
−
pi xi ȳi ,
i∈I
i∈I
i∈I
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where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) .
The following superadditivity property in terms of weights holds [2, p. 16].
Theorem 4.3. For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have
(4.6)
S (p + q, I, x, y) ≥ S (p, I, x, y) + S (q, I, x, y) ≥ 0.
Proof. Using the (CBS) −inequality for real numbers
1
1
(4.7)
a2 + b2 2 c2 + d2 2 ≥ ac + bd; a, b, c, d ≥ 0,
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
we have
! 12
X
S (p, I, x, y) =
2
pi |xi | +
i∈I
X
2
qi |xi |
! 12
X
2
pi |yi | +
i∈I
i∈I
X
qi |yi |
i∈I
Title Page
X
X
−
pi xi ȳi +
qi xi ȳi i∈I
i∈I
! 12
! 21
X
X
pi |xi |2
pi |yi |2
≥
i∈I
+
X
Contents
JJ
J
i∈I
! 12
qi |xi |2
i∈I
! 12
X
qi |yi |2
i∈I
= S (p, I, x, y) + S (q, I, x, y) ,
S.S. Dragomir
2
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X
X
−
pi xi ȳi − qi xi ȳi i∈I
i∈I
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and the inequality (4.6) is proved.
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The following corollary concerning the monotonicity of S (·, I, x, y) also
holds [2, p. 16].
Corollary 4.4. For any p, q ∈ S+ (R) with p ≥ q and I ∈ Pf (N) , x, y ∈
S (K) one has the inequality:
S (p, I, x, y) ≥ S (q, I, x, y) ≥ 0.
(4.8)
Proof. Using Theorem 4.3, we have
S (p, I, x, y) = S ((p − q) + q, I, x, y) ≥ S (p − q, I, x, y) + S (q, I, x, y)
giving
S (p, I, x, y) − S (q, I, x, y) ≥ S (p − q, I, x, y) ≥ 0
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
and the inequality (4.8) is proved.
Remark 4.2. The following inequalities follow by the above results [2, p. 17].
Title Page
1. Let αi ∈ R (i ∈ {1, . . . , n}) and xi , yi ∈ K (i ∈ {1, . . . , n}) . Then one
has the inequality:
! 21 n
n
n
X
X
X
(4.9)
|xi |2
|yi |2
−
xi ȳi i=1
i=1
i=1
! 21 n
n
n
X
X
X
≥
|xi |2 sin2 αi
|yi |2 sin2 αi
−
xi ȳi sin2 αi i=1
i=1
i=1
! 12 n
n
n
X
X
X
+
|xi |2 cos2 αi
|yi |2 cos2 αi
−
xi ȳi cos2 αi ≥ 0.
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i=1
i=1
i=1
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2. Denote Sn (1) := {p ∈ S+ (R) |pi ≤ 1 for all i ∈ {1, . . . , n}} . Then for
all x, y ∈ S (K) one has the bound (see also Corollary 4.2):
! 12 n
n
n
X
X
X
2
2
(4.10)
0≤
|xi |
|yi |
−
xi ȳi i=1
i=1
 i=1

! 12 n
n
n
 X
X

X
2
2
= sup
pi |xi |
pi |yi |
−
pi xi ȳi .

p∈Sn (1) 
i=1
4.3.
i=1
i=1
The Superadditivity as an Index Set Mapping
We assume that we are under the hypothesis and notations in Section 4.2. Reconsider the functional S (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R,
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
Title Page
! 12
(4.11)
S (p, I, x, y) :=
X
i∈I
2
pi |xi |
X
i∈I
2
pi |yi |
X
−
pi xi ȳi .
i∈I
The following superadditivity property as an index set mapping holds [2].
Theorem 4.5. For any I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅, one has the inequality
(4.12)
S (p, I ∪ J, x, y) ≥ S (p, I, x, y) + S (p, J, x, y) ≥ 0.
Proof. Using the elementary inequality for real numbers
1
1
(4.13)
a2 + b2 2 c2 + d2 2 ≥ ac + bd; a, b, c, d ≥ 0,
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we have
S (p, I ∪ J, x, y)
! 12
=
X
pi |xi |2 +
i∈I
≥
X
pj |xj |2
j∈J
! 12
X
i∈I
X
X
−
pi xi ȳi +
pj xj ȳj i∈I
j∈J
! 12
! 12
X
X
2
2
pi |yi |
+
pi |xi |
i∈I
i∈I
pi |yi |2 +
X
pj |yj |2
j∈J
! 12
X
pj |xj |2
j∈J
! 12
X
pj |yj |2
j∈J
X
X
−
pi xi ȳi − pj xj ȳj i∈I
S.S. Dragomir
Title Page
j∈J
= S (p, I, x, y) + S (p, J, x, y)
and the inequality (4.12) is proved.
The following corollary concerning the monotonicity of S (p, ·, x, y) as an
index set mapping also holds [2, p. 16].
Corollary 4.6. For any I, J ∈ Pf (N) with I ⊇ J 6= ∅, one has
(4.14)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S (p, I, x, y) ≥ S (p, J, x, y) ≥ 0.
Proof. Using Theorem 4.5, we may write
S (p, I, x, y) = S (p, (I\J) ∪ J, x, y) ≥ S (p, I\J, x, y) + S (p, J, x, y)
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giving
S (p, I, x, y) − S (p, J, x, y) ≥ S (p, I\J, x, y) ≥ 0
which proves the desired inequality (4.14).
Remark 4.3. The following inequalities follow by the above results [2, p. 17].
1. Let pi ≥ 0 (i ∈ {1, . . . , 2n}) and xi , yi ∈ K (i ∈ {1, . . . , 2n}) . Then we
have the inequality
! 12 2n
2n
2n
X
X
X
(4.15)
pi |xi |2
pi |yi |2
−
pi xi ȳi i=1
i=1
i=1
! 21 n
n
n
X
X
X
p2i xi ȳi ≥
p2i |x2i |2
p2i |y2i |2
−
i=1
i=1
i=1
! 12
n
n
X
X
+
p2i−1 |x2i−1 |2
p2i−1 |y2i−1 |2
i=1
i=1
n
X
−
p2i−1 x2i−1 ȳ2i−1 i=1
≥ 0.
S.S. Dragomir
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2. We have the bound
(4.16)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
n
X
2
pi |xi |
pi |yi |2
i=1
i=1
! 12
n
X
−
pi xi ȳi i=1
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! 12

=
sup
I⊆{1,...,n}
I6=∅
X
X
pi |xi |2
pi |yi |2
i∈I
i∈I


X
−
pi xi ȳi  ≥ 0.
i∈I
3. Define the sequence
(4.17)
Sn :=
n
X
n
X
2
pi |xi |
pi |yi |2
i=1
i=1
! 12
n
X
−
pi xi ȳi ≥ 0
i=1
where p = (pi )i∈N ∈ S+ (R) , x = (xi )i∈N , y = (yi )i∈N ∈ S (K) . Then
Sn is monotontic nondecreasing and we have the following lower bound
n
1
1
(4.18)
Sn ≥ max
pi |xi |2 + pj |xj |2 2 pi |yi |2 + pj |yj |2 2
1≤i,j≤n
− |pi xi ȳi + pj xj ȳj |
Strong Superadditivity in Terms of Weights
With the notations in Section 4.2, define the mapping
(4.19)
S.S. Dragomir
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JJ
J
≥ 0.
4.4.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
i∈I
where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) .
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2
X
X
X
S̄ (p, I, x, y) :=
pi |xi |2
pi |yi |2 − pi xi ȳi ,
i∈I
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i∈I
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Denote also by k·k`,H the weighted Euclidean norm
! 12
(4.20)
kxk`,H :=
X
2
`i |xi |
,
` ∈ S+ (R) , H ∈ Pf (N) .
i∈H
The following strong superadditivity property in terms of weights holds [2,
p. 18].
Theorem 4.7. For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have
(4.21) S̄ (p + q, I, x, y) − S̄ (p, I, x, y) − S̄ (q, I, x, y)
2


kykp,I
kxkp,I
 ≥ 0.
≥ det 
kykq,I
kxkq,I
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
Title Page
Contents
Proof. We have
(4.22)
S̄ (p + q, I, x, y)
!
!
X
X
X
X
2
2
2
2
=
pi |xi | +
qi |xi |
pi |yi | +
qi |yi |
i∈I
i∈I
i∈I
i∈I
2
X
X
−
pi xi ȳi +
qi xi ȳi i∈I
i∈I
X
X
X
X
≥
pi |xi |2
pi |yi |2 +
qi |xi |2
qi |yi |2
i∈I
i∈I
i∈I
i∈I
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!2
X
X
− pi xi ȳi + qi xi ȳi i∈I
i∈I
X
X
pi |xi |2
qi |yi |2
= S̄ (p, I, x, y) + S̄ (q, I, x, y) +
i∈I
i∈I
X
X
X
X
2
2
+
qi |xi |
pi |yi | − 2 pi xi ȳi qi xi ȳi .
i∈I
i∈I
i∈I
i∈I
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
By (CBS) −inequality, we have
"
# 12
X
X
X
X
X
X
pi xi ȳi qi xi ȳi ≤
pi |xi |2
pi |yi |2
qi |xi |2
qi |yi |2
i∈I
i∈I
i∈I
i∈I
i∈I
S.S. Dragomir
i∈I
Title Page
and thus
(4.23)
Contents
X
2
pi |xi |
i∈I
X
2
qi |yi | +
i∈I
X
2
qi |xi |
X
i∈I
2
pi |yi |

! 12
X
X
 X
2
− 2
pi xi ȳi qi xi ȳi ≥
pi |xi |
i∈I
JJ
J
i∈I
i∈I
i∈I
! 12
X
2
qi |yi |
X
i∈I
qi |xi |2
Go Back
i∈I
Close
! 12 2
! 21
−
X
II
I
pi |yi |2
i∈I
Utilising (4.22) and (4.23) we deduce the desired inequality (4.21).
 .
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The following corollary concerning a strong monotonicity result also holds
[2, p. 18].
Corollary 4.8. For any p, q ∈ S+ (R) with p ≥ q one has the inequality:
(4.24) S̄ (p, I, x, y) − S̄ (q, I, x, y)


kxkq,I
kykq,I
2
 ≥ 0.
≥ det 
kxkp−q,I
kykp−q,I
Remark 4.4. The following refinement of the (CBS) −inequality is a natural
consequence of (4.21) [2, p. 19]
2
X
X
X
(4.25)
|xi |2
|yi |2 − xi ȳi i∈I
i∈I
i∈I
2
X
X
X 2
2
2
2
2
≥
|xi | sin αi
|yi | sin αi − xi ȳi sin αi i∈I
i∈I
i∈I
2
X
X
X
+
|xi |2 cos 2 αi
|yi |2 cos 2 αi − xi ȳi cos 2 αi i∈I
i∈I
i∈I

 12 2
12
P
P
|yi |2 sin2 αi
|xi |2 sin2 αi





 i∈I
i∈I



+ det 
 ≥ 0.

 12 
12


 P
P 2
|xi |2 cos 2 αi
|yi | cos2 αi
i∈I
i∈I
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
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where αi ∈ R, i ∈ I.
4.5.
Strong Superadditivity as an Index Set Mapping
We assume that we are under the hypothesis and notations in Section 4.2. Reconsider the functional S̄ (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R,
(4.26)
S̄ (p, I, x, y) :=
X
i∈I
2
X
X
2
2
pi |xi |
pi |yi | − pi xi ȳi .
i∈I
i∈I
The following strong supperadditivity property as an index set mapping holds
[2, p. 18].
Theorem 4.9. For any p ∈ S+ (R) , I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅ and
x, y ∈ S (K) , we have
(4.27) S̄ (p, I ∪ J, x, y) − S̄ (p, I, x, y) − S̄ (p, J, x, y)


2
kxkp,I
kykp,I
 ≥ 0.
≥ det 
kxkp,J
kykp,J
S̄ (p, I ∪ J, x, y)
!
!
X
X
X
X
=
pi |xi |2 +
pj |xj |2
pi |yi |2 +
pj |yj |2
i∈I
S.S. Dragomir
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Proof. We have
(4.28)
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Type Discrete Inequalities
j∈J
i∈I
j∈J
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2
X
X
−
pi xi ȳi +
pj xj ȳj i∈I
j∈I
X
X
X
X
≥
pi |xi |2
pi |yi |2 +
pj |xj |2
pj |yj |2
i∈I
+
i∈I
X
j∈J
2
pi |xi |
i∈I
X
2
pj |yj | +
j∈J
j∈J
X
pi |yi |2
i∈I
X
pj |xj |2
j∈J
!2
X
X
− pi xi ȳi + pj xj ȳj i∈I
j∈I
X
X
pi |xi |2
pj |yj |2
= S̄ (p, I, x, y) + S̄ (p, J, x, y) +
i∈I
j∈J
i∈I
j∈I
By the (CBS) −inequality, we have
X
X
pi xi ȳi pj xj ȳj i∈I
j∈I
"
# 12
X
X
X
X
≤
pi |xi |2
pi |yi |2
pj |xj |2
pj |yj |2
i∈I
i∈I
j∈J
S.S. Dragomir
j∈J
X
X
X
X
2
2
+
pi |yi |
pj |xj | − 2 pi xi ȳi pj xj ȳj .
i∈I
A Survey on
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Type Discrete Inequalities
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and thus
(4.29)
X
pi |xi |2
i∈I
X
pj |yj |2 +
j∈J
X
pi |yi |2
X
i∈I
pj |xj |2
j∈J

! 21
X
X
 X
2
pi |xi |
− 2
pi xi ȳi pj xj ȳj ≥
i∈I
j∈I
i∈I
! 12
X
2
pj |yj |
j∈J
! 12 2
! 12
−
X
pi |yi |2
X
i∈I
pj |xj |2
 .
j∈J
If we use now (4.28) and (4.29), we may deduce the desired inequality (4.27).
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
Title Page
The following corollary concerning strong monotonicity also holds [2, p.
18].
Corollary 4.10. For any I, J ∈ Pf (N) \ {∅} with I ⊇ J one has the inequality
(4.30)
S̄ (p, I, x, y) − S̄ (p, J, x, y)


kxkp,J


≥ det
kxkp,I\J
kykp,J
2
 ≥ 0.
kykp,I\J
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Remark 4.5. The following refinement of the (CBS) −inequality is a natural
consequence of (4.27) [2, p. 19].
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Suppose pi ≥ 0, i ∈ {1, . . . , 2n} and xi , yi ∈ K, i ∈ {1, . . . , 2n} . Then we
have the inequality
(4.31)
2n
X
i=1
2
2n
2n
X
X
pi |xi |2
pi |yi |2 − pi xi ȳi i=1
i=1
2
n
X
2
2
p2i |x2i |
p2i |y2i | − p2i x2i ȳ2i ≥
i=1
i=1
i=1
2
n
n
n
X
X
X
+
p2i−1 |x2i−1 |2
p2i−1 |y2i−1 |2 − p2i−1 x2i−1 ȳ2i−1 i=1
i=1
i=1
2

 n
12
21
n
P
P
2
2
p2i |y2i |
p2i |x2i |





 i=1
i=1



+ det 
 ≥ 0.

 1
1 
n
n
2
2 

 P
P
p2i−1 |x2i−1 |2
p2i−1 |y2i−1 |2
n
X
i=1
4.6.
n
X
i=1
Another Superadditivity Property
Let Pf (N) be the family of finite parts of the set of natural numbers, S (R)
the linear space of real sequences and S+ (R) the family of nonnegative real
sequences.
A Survey on
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S.S. Dragomir
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Consider the mapping C : S+ (R) × Pf (N) × S (R) × S (R) → R
!2
X
X
X
(4.32)
C p, I, a, b :=
pi a2i
pi b2i −
p i ai b i .
i∈I
i∈I
i∈I
The following identity holds [3, p. 115].
Lemma 4.11. For any p, q ∈ S+ (R) one has
(4.33) C p + q, I, a, b = C p, I, a, b + C q, I, a, b
X
+
pi qj (ai bj − aj bi )2 .
(i,j)∈I×I
S.S. Dragomir
Proof. Using the well-known Lagrange’s identity, we have
1 X
(4.34)
C p, I, a, b =
pi pj (ai bj − aj bi )2 .
2
Title Page
Contents
(i,j)∈I×I
Thus
JJ
J
C p + q, I, a, b
1 X
=
(pi + qi ) (pj + qj ) (ai bj − aj bi )2
2
X
(i,j)∈I×I
1
+
2
1
pi pj (ai bj − aj bi ) +
2
2
X
(i,j)∈I×I
X
Close
2
qi qj (ai bj − aj bi )
Quit
(i,j)∈I×I
1
pi qj (ai bj − aj bi ) +
2
2
X
(i,j)∈I×I
II
I
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(i,j)∈I×I
1
=
2
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Type Discrete Inequalities
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2
pj qi (ai bj − aj bi )
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= C p, I, a, b + C q, I, a, b +
X
pi qj (ai bj − aj bi )2
(i,j)∈I×I
since, by symmetry,
X
pi qj (ai bj − aj bi )2 =
(i,j)∈I×I
X
pj qi (ai bj − aj bi )2 .
(i,j)∈I×I
Consider the following mapping:

X
X
12
D p, I, a, b := C p, I, a, b
=
pi a2i
pi b2i −
i∈I
i∈I
!2  21
X
p i ai b i  .
S.S. Dragomir
i∈I
The following result has been obtained in [4, p. 88] as a particular case of a
more general result holding in inner product spaces.
Theorem 4.12. For any p, q ∈ S+ (R) , I ∈ Pf (N) and a, b ∈ S (R) , we have
the superadditive property
(4.35)
D p + q, I, a, b ≥ D p, I, a, b + D q, I, a, b ≥ 0.
Proof. We will give here an elementary proof following the one in [3, p. 116 –
p. 117].
By Lemma 4.11, we obviously have
(4.36) D2 p + q, I, a, b = D2 p, I, a, b + D2 q, I, a, b
X
+
pi qj (ai bj − aj bi )2 .
(i,j)∈I×I
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Type Discrete Inequalities
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We claim that
X
(4.37)
pi qj (ai bj − aj bi )2 ≥ 2D p, I, a, b D q, I, a, b .
(i,j)∈I×I
Taking the square in both sides of (4.37), we must prove that
#2
"
(4.38)
X
pi a2i
i∈I
X
qi b2i +
i∈I
X
qi a2i
X
i∈I
pi b2i −2
i∈I
p i ai b i
X
i∈I

X
X
≥ 4
pi a2i
pi b2i −
i∈I
X
i∈I
q i ai b i
i∈I
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
!2 
X
p i ai b i 
S.S. Dragomir
i∈I

X
X
qi b2i −
×
qi a2i
i∈I
i∈I
!2 
X
q i ai b i  .
i∈I
Contents
Let us denote
! 21
a :=
X
pi a2i
! 12
,
x :=
X
i∈I
qi a2i
! 12
,
b :=
i∈I
X
pi b2i
i∈I
! 21
y :=
X
i∈I
qi b2i
,
c :=
X
i∈I
p i ai b i ,
z :=
X
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,
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q i ai b i .
i∈I
With these notations (4.38) may be written in the following form
2
(4.39)
a2 y 2 + b2 x2 − 2cz ≥ 4 a2 b2 − c2 x2 y 2 − z 2 .
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Using the elementary inequality
m2 − n2 p2 − q 2 ≤ (mp − nq)2 , m, n, p, q ∈ R
we may state that
4 (abxy − cz)2 ≥ 4 a2 b2 − c2
(4.40)
x2 y 2 − z 2 ≥ 0.
Since, by the (CBS) −inequality, we observe that abxy ≥ |cz| ≥ |cz| , we can
state that
a2 y 2 + b2 x2 − 2cz ≥ 2 (abxy − cz) ≥ 0
giving
a2 y 2 + b2 x2 − 2cz
(4.41)
2
≥ 4 (abxy − cz)2 .
S.S. Dragomir
Utilizing (4.40) and (4.41) we deduce the inequality (4.39), and (4.37) is proved.
Finally, by (4.36) and (4.37) we have
2
D2 p + q, I, a, b ≥ D p, I, a, b + D q, I, a, b ,
i.e., the superadditivity property (4.35).
Remark 4.6. The following refinement of the (CBS) − inequality holds [4, p.
89]

!2  21
X X
X
(4.42) 
a2i
b2i −
ai b i 
i∈I
i∈I
i∈I
i∈I
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i∈I

X
X
≥
a2i sin2 αi
b2i sin2 αi −
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Type Discrete Inequalities
!2  12
X
i∈I
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2
ai bi sin αi

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
X
X
+
a2i cos 2 αi
b2i cos2 αi −
i∈I
X
i∈I
!2  21
ai bi cos2 αi  ≥ 0
i∈I
for any αi ∈ R, i ∈ {1, . . . , n} .
4.7.
The Case of Index Set Mapping
Assume that we are under the hypothesis and notations in Section 4.6. Reconsider the functional C : S+ (R) × Pf (N) × S (R) × S (R) → R given by
!2
(4.43)
C p, I, a, b :=
X
pi a2i
i∈I
X
pi b2i
−
i∈I
X
p i ai b i
S.S. Dragomir
.
i∈I
Title Page
The following identity holds.
Contents
Lemma 4.13. For any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ one has the identity:
(4.44) C p, I ∪ J, a, b
= C p, I, a, b + C p, J, a, b +
X
pi pj (ai bj − aj bi )2 .
(i,j)∈I×J
Proof. Using Lagrange’s identity [5, p. 84], we may state
1
(4.45) C p, K, a, b =
2
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X
(i,j)∈K×K
2
pi pj (ai bj − aj bi ) , K ∈ Pf (N) \ {∅} .
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Thus
C p, I ∪ J, a, b
X
1
=
2
pi pj (ai bj − aj bi )2
(i,j)∈(I∪J)×(I∪J)
=
1
2
X
pi pj (ai bj − aj bi )2 +
(i,j)∈I×I
+
1
2
X
1
2
X
pi pj (ai bj − aj bi )2
(i,j)∈I×J
1 X
pi pj (ai bj − aj bi )2
2
(i,j)∈J×J
X
pi pj (ai bj − aj bi )2
A Survey on
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Type Discrete Inequalities
pi pj (ai bj − aj bi )2 +
(i,j)∈J×I
= C p, I, a, b + C p, J, a, b +
S.S. Dragomir
(i,j)∈I×J
since, by symmetry,
X
pi pj (ai bj − aj bi )2 =
(i,j)∈I×J
Title Page
X
Contents
pi pj (ai bj − aj bi )2 .
JJ
J
(i,j)∈J×I
II
I
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Now, if we consider the mapping
Close

X
X
12
=
pi a2i
pi b2i −
D p, I, a, b := C p, I, a, b
i∈I
i∈I
!2  12
X
p i ai b i
i∈I
then the following superadditivity property as an index set mapping holds:
 ,
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Theorem 4.14. For any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ one has
(4.46)
D p, I ∪ J, a, b ≥ D p, I, a, b + D p, J, a, b ≥ 0.
Proof. By Lemma 4.13, we have
(4.47) D2 p, I ∪ J, a, b
X
= D2 p, I, a, b + D2 p, J, a, b +
pi pj (ai bj − aj bi )2
A Survey on
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Type Discrete Inequalities
(i,j)∈I×J
To prove (4.46) it is sufficient to show that
X
(4.48)
pi pj (ai bj − aj bi )2 ≥ 2D p, I, a, b D p, J, a, b .
S.S. Dragomir
(i,j)∈I×J
Title Page
Taking the square in (4.48), we must demonstrate that
Contents
JJ
J
#2
"
X
i∈I
pi a2i
X
j∈J
pj b2j
+
X
j∈J
pj a2j
X
pi b2i
−2
i∈I
p i ai b i
X
i∈I

X
X
≥ 4
pi a2i
pi b2i −
i∈I
X
i∈I
p j aj b j
j∈J
Go Back
!2 
X
p i ai b i
Close

i∈I

X
X
×
pj a2j
pj b2j −
j∈J
II
I
j∈J
Quit
!2 
X
j∈J
p j aj b j
.
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If we denote
! 12
! 12
a :=
X
pi a2i
,
X
x :=
i∈I
pj a2j
! 21
,
X
b :=
j∈J
pi b2i
,
i∈I
! 21
y :=
X
pj b2j
,
c :=
j∈J
X
p i ai b i ,
z :=
i∈I
X
p j aj b j ,
j∈J
then we need to prove
(4.49)
a2 y 2 + b2 x2 − 2cz
2
≥ 4 a2 b 2 − c 2
A Survey on
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x2 y 2 − z 2 ,
S.S. Dragomir
which has been shown in Section 4.6.
This completes the proof.
Remark 4.7. The following refinement of the (CBS) −inequality holds
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
2n
2n
X
X
2

p i ai
pi b2i −
i=1
i=1
2n
X
!2  12
p i ai b i
i=1

n
n
X
X
2

≥
p2i a2i
p2i b22i −
i=1
i=1

n
n
X
X
2

+
p2i−1 a2i−1
p2i−1 b22i−1 −
i=1
JJ
J

i=1
n
X
!2  12
p2i a2i b2i
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
i=1
n
X
i=1
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I
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!2  21
p2i−1 a2i−1 b2i−1
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4.8.
Supermultiplicity in Terms of Weights
Denote by S+ (R) the set of nonnegative sequences. Assume that A : S+ (R) →
R is additive on S+ (R) , i.e.,
(4.50)
A (p + q) = A (p) + A (q) ,
p, q ∈ S+ (R)
and L : S+ (R) → R is superadditive on S+ (R) , i.e.,
(4.51)
L (p + q) ≥ L (p) + L (q) , p, q ∈ S+ (R) .
Define the following associated functionals
(4.52)
L (p)
and H (p) := [F (p)]A(p) .
F (p) :=
A (p)
A Survey on
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S.S. Dragomir
Title Page
The following result holds [3, Theorem 2.1].
Contents
Lemma 4.15. With the above assumptions, we have
(4.53)
H (p + q) ≥ H (p) H (q) ;
for any p, q ∈ S+ (R) , i.e., H (·) is supermultiplicative on S+ (R) .
Proof. We shall follow the proof in [3].
Using the well-known arithmetic mean-geometric mean inequality for real
numbers
(4.54)
α
αx + βy
≥ x α+β y
α+β
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β
α+β
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for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively
(4.55)
L (p + q)
A (p + q)
L (p + q)
=
A (p) + A (q)
L (p) + L (q)
≥
A (p) + A (q)
F (p + q) =
=
L(p)
L(q)
A (p) A(p)
+ A (q) A(q)
A (p) + A (q)
A (p) F (p) + A (q) F (q)
=
A (p) + A (q)
A(p)
A(q)
≥ [F (p)] A(p)+A(q) · [F (q)] A(p)+A(q)
for all p, q ∈ S+ (R) . However, A (p) + A (q) = A (p + q) , and thus (4.55)
implies the desired inequality (4.53).
We are now able to point out the following inequality related to the (CBS) −
inequality.
The first result is incorporated in the following theorem [3, p. 115].
Theorem 4.16. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality
(
"
X
X
1
(4.56)
(pi + qi ) a2i
(pi + qi ) b2i
PI + QI i∈I
i∈I
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S.S. Dragomir
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−
X
i∈I
!2 PI +QI

(pi + qi ) ai bi 

 
!2 PI
1 X

X
X
2
2


≥
pi ai
p i bi −
p i ai b i
 PI

i∈I
i∈I
i∈I


!2 QI

 1 X
X
X
2
2


> 0,
q i ai
qi b i −
q i ai b i
×

 QI
i∈I
i∈I
i∈I
where PI :=
P
i∈I
pi > 0, QI :=
P
i∈I
qi > 0.
S.S. Dragomir
Proof. Consider the functionals
X
A (p) :=
p i = PI ;
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i∈I
!2
C (p) :=
X
i∈I
pi a2i
X
i∈I
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Type Discrete Inequalities
pi b2i
−
X
pi ai bi
.
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Then A (·) is additive and C (·) is superadditive (see for example Lemma 4.11)
on S+ (R) .
Applying Lemma 4.15 we deduce the desired inequality (4.56).
The following refinement of the (CBS) −inequality holds.
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Corollary 4.17. For any a, b, α ∈ S (R) , one has the inequality
(4.57)
n
X
a2i
i=1
n
X
b2i
n
X
−
i=1
!2
ai b i
i=1
1
≥
1
n
Pn
n1 Pni=1 sin2 αi
2
i=1 sin αi

n
n
X
X
2
2

×
ai sin αi
b2i sin2 αi −
i=1
Pn
2
i=1 cos αi
n
X
i=1
n1 Pni=1 cos 2 αi
!2  n1
ai bi sin2 αi 
Pn
i=1
sin2 αi
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
i=1

n
n
X
X
2
2

×
ai cos αi
b2i cos2 αi −
i=1
1
n
n
X
!2  n1
ai bi cos2 αi
Pn
i=1
S.S. Dragomir
cos2 αi
≥ 0.

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i=1
i=1
Contents
The following result holds [3, p. 116].
Theorem 4.18. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality
(4.58)
"

1
 PI + QI
# 12 "
X
i∈I
(pi + qi ) a2i
1
PI + QI
# 12
X
(pi + qi ) b2i
i∈I
)PI +QI
X
1
−
(pi + qi ) ai bi PI + QI
i∈I
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
1 X 2
p i ai
PI i∈I
≥
! 12

1 X 2
×
q i ai
QI i∈I
PI
1 X 2
1 X
p i bi
−
p i ai b i 
PI
PI i∈I
i∈I
QI
! 12 X
X
1
1
2
qi b i
−
qi ai bi  ≥ 0.
QI
QI
! 12
! 12
i∈I
i∈I
Proof. Follows by Lemma 4.15 on taking into account that the functional
! 12
! 12 X
X
X
−
p i ai b i pi a2i
pi b2i
B (p) :=
i∈I
i∈I
i∈I
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S.S. Dragomir
is superadditive on S+ (R) (see Section 4.2).
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The following refinement of the (CBS) −inequality holds.
Corollary 4.19. For any a, b, α ∈ S (R) , one has the inequality
n
X
(4.59)
!
1
2
1
2
!
n
n
X
X
2
bi
−
ai b i a2i
i=1
1
≥
1
n

×
n
X
i=1
i=1
Pn
a2i
2
i=1
2
sin αi
! 12
sin αi
1
1
n
Pn
II
I
i=1
cos2 αi
Close
n1 Pni=1 cos2 αi
P
 n1 ni=1 sin2 αi
! 12 n
n
X
X
b2i sin2 αi
−
ai bi sin2 αi 
i=1
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i=1
n1 Pni=1 sin2 αi ·
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
×
n
X
! 12
n
X
a2i cos2 αi
i=1
! 12
b2i cos2 αi
n
 n1
X
2
−
ai bi cos αi 
Pn
i=1
cos2 αi
≥ 0.
i=1
i=1
Finally, we may also state [3, p. 117].
Theorem 4.20. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality
"
X
X
1
1
(4.60)
(pi + qi ) a2i ·
(pi + qi ) b2i
PI + QI i∈I
PI + QI i∈I
I
!2  PI +Q
2
X
1
−
(pi + qi ) ai bi 
PI + QI i∈I

≥
1 X 2 1 X 2
p i ai ·
p i bi −
PI i∈I
PI i∈I
1 X
p i ai b i
PI i∈I
!2  P2I
1 X 2 1 X 2
×
q i ai ·
qi b i −
QI i∈I
QI i∈I
1 X
q i ai b i
QI i∈I
i=1
i=1
Title Page
JJ
J
!2  Q2I

Proof. Follows by Lemma 4.15 on taking into account that the functional

!2  21
n
n
n
X
X
X
pi a2i
pi b2i −
p i ai b i 
D (p) := 
i=1
S.S. Dragomir
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

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is superadditive on S+ (R) (see Section 4.6).
The following corollary also holds.
Corollary 4.21. For any a, b, α ∈ S (R) , one has the inequality
(4.61)

n
n
X
X

a2i
b2i −
i=1
i=1
1
n
!2  12
ai b i 
i=1
1
≥
n
X
Pn
2
i=1 sin αi
n1 Pni=1 sin2 αi ·

n
n
X
X
×
a2i sin2 αi
b2i sin2 αi −
i=1
i=1

n
n
X
X
2
2

×
ai cos αi
b2i cos2 αi −
i=1
4.9.
i=1
1
1 Pn
2
2 α n i=1 cos αi
cos
i
i=1
 2n1 Pni=1 sin2 αi
!
2
n
X
ai bi sin2 αi 
1
n
Pn
i=1
n
X
!2  2n1
ai bi cos2 αi
Pn

i=1
S.S. Dragomir
Title Page
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cos2 αi
≥ 0.
i=1
Supermultiplicity as an Index Set Mapping
Denote by Pf (N) the set of all finite parts of the natural number set N and
assume that B : Pf (N) → R is set-additive on Pf (N) , i.e.,
(4.62)
A Survey on
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B (I ∪ J) = B (I) + B (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅,
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and G : Pf (N) → R is set-superadditive on Pf (N) , i.e.,
(4.63)
G (I ∪ J) ≥ G (I) + G (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅.
We may define the following associated functionals
(4.64)
M (I) :=
G (I)
and N (I) := [M (I)]A(I) .
B (I)
With these notations we may prove the following lemma that is interesting
in itself as well.
Lemma 4.22. Under the above assumptions one has
(4.65)
N (I ∪ J) ≥ N (I) N (J)
for any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅, i.e., N (·) is set-supermultiplicative
on Pf (N) .
Proof. Using the arithmetic mean – geometric mean inequality
β
α
αx + βy
≥ x α+β y α+β
α+β
(4.66)
for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively for
I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ that
(4.67)
G (I ∪ J)
B (I ∪ J)
G (I ∪ J)
=
B (I) + B (J)
M (I ∪ J) =
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≥
=
G (I) + G (J)
B (I) + B (J)
G(I)
G(J)
B (I) B(I)
+ B (J) B(J)
B (I) + B (J)
B (I) M (I) + B (J) M (J)
=
B (I) + B (J)
B(I)
B(J)
≥ (M (I)) B(I)+B(J) · (M (J)) B(I)+B(J) .
Since B (I) + B (J) = B (I ∪ J) , we deduce by (4.67) the desired inequality
(4.65).
Now, we are able to point out some set-superadditivity properties for some
functionals associates to the (CBS) −inequality.
The first result is embodied in the following theorem.
Theorem 4.23. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that
I ∩ J 6= ∅, then one has the inequality
(4.68)


!2 PI∪J
 1

X
X
X
2
2


p k ak
p k bk −
p k ak b k
 PI∪J

k∈I∪J
k∈I∪J
k∈I∪J
 
!2 PI
1 X

X
X

≥
pi a2i
pi b2i −
p i ai b i 
 PI

i∈I
i∈I
i∈I
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 
 1 X
X

×
pj a2j
pj b2j −
 PJ
j∈J
j∈J
when PJ :=
P
j∈J
X
j∈J
!2 PJ

,
p j aj b j 

pj .
Proof. Consider the functionals
X
B (I) :=
pi ;
i∈I
!2
G (I) :=
X
i∈I
pi a2i
X
i∈I
pi b2i −
X
p i ai b i
.
i∈I
The functional B (·) is obviously set-additive and (see Section 4.7) the functional G (·) is set-superadditive. Applying Lemma 4.22 we then deduce the
desired inequality (4.68).
The following corollary is a natural application.
Corollary 4.24. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has
the inequality


!2 P2n
2n
2n
2n
 1 X

X
X

(4.69)
pi a2i
pi b2i −
p i ai b i 
 P2n

i=1
i=1
i=1
P


!2  ni=1 p2i
n
n
n


X
X
X
1

≥ Pn
p2i a22i
p2i b22i −
p2i a2i b2i 
 i=1 p2i

i=1
i=1
i=1
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(
×
" n
n
X
X
1
2
Pn
p2i−1 a2i−1
p2i−1 b22i−1
p
2i−1
i=1
i=1
i=1
P
!2  ni=1 p2i−1
n

X
−
p2i−1 a2i−1 b2i−1 
.

i=1
The following result also holds.
Theorem 4.25. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that
I ∩ J 6= ∅, then one has the inequality
"
)PI∪J
# 12 # 12 "
1 X
 1 X
X
1
p k ak b k pk b2k − (4.70)
pk a2k
PI∪J
 PI∪J
PI∪J k∈I∪J
k∈I∪J
k∈I∪J

PI
! 12 ! 12

 1 X
1 X 2
1 X
−
p i ai b i ≥
p i bi
pi a2i

 PI
P
P
I
I
i∈I
i∈I
i∈I

PJ
! 12
! 12 
 1 X
X
X
1
1
2
2
×
p j aj
p j bj
−
p j aj b j .
 PJ
PJ

PJ
j∈J
j∈J
j∈J
Proof. Follows by Lemma 4.22 on taking into account that the functional
! 12
! 12 X
X
X
G (I) :=
pi a2i
pi b2i
−
p i ai b i i∈I
is set-superadditive on Pf (N) .
i∈I
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The following corollary is a natural application.
Corollary 4.26. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has
the inequality

(4.71)
2n
X
 1
pi a2i
P2n i=1

1
≥  Pn
! 12
2n
1 X 2
p i bi
P2n i=1
n
X
! 12
P2n
2n
1 X
−
p i ai b i 
P2n
i=1
! 12
p2i a22i
1
Pn
n
X
! 12
p2i b22i
p2i i=1
i=1 p2i i=1
#Pni=1 p2i
n
X
1
− Pn
p2i a2i b2i i=1 p2i
i=1

! 12
! 12
n
n
X
X
1
1
Pn
×  Pn
p2i−1 a22i−1
p2i−1 b22i−1
p
p
i=1 2i−1 i=1
i=1 2i−1 i=1
#Pni=1 p2i−1
n
X
P 1
− n
p2i−1 a2i−1 b2i−1 .
i=1 p2i−1
i=1
i=1
Finally, we may also state:
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Theorem 4.27. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that
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I ∩ J 6= ∅, then one has the inequality
!2  PI∪J
2

(4.72)
X
1 X
 1
pk a2k ·
pk b2k −
PI∪J k∈I∪J
PI∪J k∈I∪J
1
PI∪J
X
p k ak b k
k∈I∪J

≥
1 X 2 1 X 2
p i ai ·
p i bi −
PI i∈I
PI i∈I
1 X
p i ai b i
PI i∈I
!2  P2I


×
1 X
1 X 2
pj a2j ·
p j bj −
PJ j∈J
PJ j∈J

1 X
p j aj b j
PJ j∈J
A Survey on
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Type Discrete Inequalities
!2  P2J

.
Proof. Follows by Lemma 4.22 on taking into account that the functional

!2  12
X
X
X
Q (I) := 
pi a2i
pi b2i −
p i ai b i 
i∈I
i∈I
i∈I
is set-superadditive on Pf (N) (see Section 4.7).
Corollary 4.28. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has
the inequality
(4.73)
2n
2n
X
1 X 2
2
 1
p i ai ·
p i bi −
P2n i=1
P2n i=1
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The following corollary holds as well.
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2n
1 X
p i ai b i
P2n i=1
!2  P2n
2
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
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"
≥
Pn
i=1 p2i
−
"
×
n
X
1
1
i=1 p2i−1
!
p2i a22i
n
X
Pn
i=1 p2i
n
X
Pn
2
p2i a2i b2i

i=1
!
p2i−1 a22i−1
i=1
−
p2i b22i
p
2i
i=1
i=1
1 Pn

! 2 i=1 p2i
Pn
i=1
1
!
n
X
1
1
i=1 p2i−1
n
X
Pn
!
p2i−1 b22i−1
i=1
n
X
1
p2i−1 a2i−1 b2i−1
i=1 p2i−1 i=1
Pn
!2  12

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Pn
i=1 p2i−1
.
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References
[1] S.S. DRAGOMIR AND Š.Z. ARSLANAGIĆ, The improvement of CauchyBuniakowski-Schwartz’s inequality, Mat. Bilten, 16 (1992), 77–80.
[2] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity
of Schwartz’s inequality in inner product spaces, Contributions, Ser. Math.
Sci. Macedonian Acad. Sci., 15(1) (1994), 5–22.
[3] R.P. AGARWAL AND S.S. DRAGOMIR, The property of supermultiplicity
for some classical inequalities and applications, Computer Math. Applic.,
35(6) (1998), 105–118.
[4] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity
of Gram’s inequality and related results, Acta Math. Hungarica, 71(1-2)
(1996), 75–90.
[5] D.S. MITRINOVIĆ, J.E. PEC̆ARIĆ AND A.M. FINK, Classical and
New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
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S.S. Dragomir
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5.
5.1.
Reverse Inequalities
The Cassels’ Inequality
The following result was proved by J.W.S. Cassels in 1951 (see Appendix 1 of
[2] or Appendix of [3]):
Theorem 5.1. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be sequences of positive
real numbers and w̄ = (w1 , . . . , wn ) a sequence of nonnegative real numbers.
Suppose that
ai
ai
.
and M = max
(5.1)
m = min
bi
bi
i=1,n
i=1,n
Then one has the inequality
Pn
Pn
2
2
(m + M )2
i=1 wi ai
i=1 wi bi
(5.2)
≤
.
P
2
4mM
( ni=1 wi ai bi )
The equality holds in (5.2) when w1 =
m = abn1 and M = ban1 .
1
,
a1 b1
wn =
1
,
an bn
w2 = · · · = wn−1 = 0,
Proof. 1. The original proof by Cassels (1951) is of interest. We shall follow
the paper [5] in sketching this proof.
We begin with the assertion that
(5.3)
(1 + kw) (1 + k −1 w)
(1 + k) (1 + k −1 )
, k > 0, w ≥ 0
≤
4
(1 + w)2
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which, being an equivalent form of (5.2) for n = 2, shows that it holds for
n = 2.
To prove that the maximum of (5.2) is obtained when we have more than two
wi ’s being nonzero, Cassels then notes that if for example, w1 , w2 , w3 6= 0 lead
, then we would have the linear equations
to an extremum M of XY
22
a2n X + b2n Y − 2M an bn Z = 0, k = 1, 2, 3.
Nontrivial solutions exist if and only if the three vectors [a2n , b2n , an bn ] are linearly dependent. But this will be so only if, for some i 6= j (i, j = 1, 2, 3)
ai = γaj , bi = γbj . And if that were true, we could, for example, drop the ai ,
bi terms and so deal with the same problem with one less variable. If only one
wi 6= 0, then M = 1, the lower bound. So we need only examine all pairs
wi 6= 0, wj 6= 0. The result (5.2) then quickly follows.
2. We will now use the barycentric method of Frucht [1] and Watson [4].
We will follow the paper [5].
We substitute wi = ub2i in the left hand side of (5.2), which may then be
i
expressed as the ratio
N
D2
where
n 2
n X
X
ai
ai
ui and D =
ui ,
N=
b
b
i
i
i=1
i=1
Pn
assuming without loss of generality, that i=1 ai = 1. But the pointwith co
a2
ordinates (D, N ) must lie within the convex closure of the n points abii , b2i .
i
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n o
The value of DN2 at points on the parabola is one unit. If m = min abii and
i=1,n
n o
ai
M = max bi , then the minimum must lie on the chord joining the point
i=1,n
(m, m2 ) and (M, M 2 ) . Some easy calculus then leads to (5.2).
The following “unweighted” Cassels’ inequality holds.
Corollary 5.2. If ā and b̄ satisfy the assumptions in Theorem 5.1, one has the
inequality
Pn 2 Pn 2
(m + M )2
i=1 ai
i=1 bi
(5.4)
≤
.
Pn
2
4mM
( i=1 ai bi )
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The following two additive versions of Cassels inequality hold.
Corollary 5.3. With the assumptions of Theorem 5.1, one has
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(5.5)
0≤
n
X
wi a2i
i=1
√
≤
n
X
! 12
wi b2i
−
n
X
i=1
wi ai bi
i=1
√ 2
n
M− m X
√
wi ai bi .
2 mM
i=1
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and
(5.6)
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J
0≤
n
X
i=1
wi a2i
n
X
i=1
wi b2i
−
n
X
i=1
!2
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(M − m)2
≤
4mM
n
X
!2
wi ai bi
.
i=1
Proof. Taking the square root in (5.2) we get
1
P
P
( ni=1 wi a2i ni=1 wi b2i ) 2
M +m
Pn
≤ √
.
1≤
2 mM
i=1 wi ai bi
Subtracting 1 on both sides, a simple calculation will lead to (5.5).
The second inequality follows by (5.2) on subtracting 1 and appropriate computation.
A Survey on
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S.S. Dragomir
The following additive version of unweighted Cassels inequality also holds.
Corollary 5.4. With the assumption of Theorem 5.1 for ā and b̄ one has the
inequalities
√
√ 2
! 21
n
n
n
n
M− m X
X
X
X
2
2
√
(5.7)
0≤
ai
bi
−
ai b i ≤
ai b i
2
mM
i=1
i=1
i=1
i=1
and
(5.8)
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0≤
n
X
i=1
a2i
n
X
i=1
b2i −
n
X
i=1
!2
ai b i
≤
(M − m)2
4mM
n
X
i=1
!2
ai b i
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5.2.
The Pólya-Szegö Inequality
The following inequality was proved in 1925 by Pólya and Szegö [6, pp. 57,
213 – 214], [7, pp. 71– 72, 253 – 255].
Theorem 5.5. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
positive real numbers. If
(5.9) 0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for each i ∈ {1, . . . , n} ,
then one has the inequality
Pn 2 Pn 2
(ab + AB)2
i=1 ai
i=1 bi
(5.10)
≤
.
P
2
4abAB
( ni=1 ai bi )
The equality holds in (5.10) if and only if
A
A B
B
A B
p=n·
+
and q = n ·
+
a
a
b
b
a
b
are integers and if p of the numbers a1 , . . . , an are equal to a and q of these
numbers are equal to A, and if the corresponding numbers bi are equal to B
and b respectively.
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Proof. Following [5], we shall present here the original proof of Pólya and
Szegö.
We may, without loss of generality, suppose that a1 ≥ · · · ≥ an , then to
maximise the left-hand side of (5.10) we must have that the critical bi ’s be reversely ordered (for if bk > bm with k < m, then we can interchange bk and
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bm such that b2k + b2m = b2m + b2k and ak bk + am bm ≥ ak bm + am bk ), i.e., that
b1 ≤ · · · ≤ bn .
Pólya and Szegö then continue by defining nonnegative numbers ui and vi
for i = 1, . . . , n − 1 and n > 2 such that
a2i = ui a21 + vi a2n and b2i = ui b21 + vi b2n .
(5.11)
Since ai bi > ui a1 b1 + vi an bn the left hand side of (5.10),
Pn 2 Pn 2
(U a21 + V a2n ) (U b21 + V b2n )
i=1 ai
i=1 bi
≤
,
Pn
2
(U a1 b1 + V an bn )2
( i=1 ai bi )
P
P
where U = ni=1 ui and V = ni=1 vi .
This reduces the problem to that with n = 2, which is solvable by elementary
methods, leading to
Pn 2 Pn 2
(a1 b1 + an bn )2
i=1 ai
i=1 bi
(5.12)
≤
,
P
2
4a1 an b1 bn
( ni=1 ai bi )
where, since the ai ’s and bi ’s here are reversely ordered,
(5.13)
a1 = max {ai } , an = min {ai } , b1 = min {bi } , bn = max {bi } .
i=1,n
i=1,n
i=1,n
i=1,n
If we now assume, as in (5.9), that
0 < a ≤ ai ≤ A, 0 < b ≤ bi ≤ B, i = (1, . . . , n) ,
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then
(because
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(a1 b1 + an bn )2
(ab + AB)2
≤
4a1 an b1 bn
4abAB
(k+1)2
4k
≤
(α+1)2
4α
for k ≤ α), and the inequality (5.10) is proved.
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Remark 5.1. The inequality (5.10) may also be obtained from the “unweighted”
Cassels’ inequality
Pn 2 Pn 2
(m + M )2
i=1 ai
i=1 bi
(5.14)
≤
,
Pn
2
4mM
( i=1 ai bi )
where 0 < m ≤
ai
bi
≤ M for each i ∈ {1, . . . , n} .
The following additive versions of the Pólya-Szegö inequality also hold.
Corollary 5.6. With the assumptions in Theorem 5.5, one has the inequality
√
√ 2
! 21
n
n
n
n
AB − ab X
X
X
X
√
−
ai b i ≤
(5.15)
0≤
a2i
b2i
ai b i
2
abAB
i=1
i=1
i=1
i=1
and
(5.16)
0≤
n
X
i=1
5.3.
a2i
n
X
i=1
b2i −
n
X
i=1
!2
ai b i
2
≤
(AB − ab)
4abAB
n
X
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!2
ai bi
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.
i=1
The Greub-Rheinboldt Inequality
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The following weighted version of the Pólya-Szegö inequality was obtained by
Greub and Rheinboldt in 1959, [6].
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Theorem 5.7. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
positive real numbers and w̄ = (w1 , . . . , wn ) a sequence of nonnegative real
numbers. Suppose that
Quit
(5.17)
0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ (i = 1, . . . , n) .
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Then one has the inequality
Pn
Pn
2
2
(ab + AB)2
i=1 wi ai
i=1 wi bi
(5.18)
≤
.
P
2
4abAB
( ni=1 wi ai bi )
Equality holds in (5.18) when wi = a11b1 , wn = an1bn , w2 = · · · = wn−1 = 0,
m = abn1 , M = ban1 with a1 = A, an = a, b1 = b and bn = b.
Remark 5.2. This inequality follows by Cassels’ result which states that
Pn
Pn
2
2
w
a
(m + M )2
i
i
i=1
i=1 wi bi
(5.19)
≤
,
P
2
4mM
( ni=1 wi ai bi )
provided 0 < m ≤
ai
bi
≤ M < ∞ for each i ∈ {1, . . . , n} .
The following additive versions of Greub-Rheinboldt also hold.
Corollary 5.8. With the assumptions in Theorem 5.7, one has the inequalities
! 12
n
n
n
X
X
X
(5.20)
0≤
wi a2i
wi b2i
−
wi ai bi
i=1
√
≤
i=1
i=1
√ 2
n
AB − ab X
√
wi ai bi
2 abAB
i=1
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(5.21)
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0≤
n
X
i=1
wi a2i
n
X
i=1
wi b2i
−
n
X
i=1
!2
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(AB − ab)2
≤
4abAB
5.4.
n
X
!2
wi ai bi
.
i=1
A Cassels’ Type Inequality for Complex Numbers
The following reverse inequality for the (CBS) −inequality holds [9].
Theorem 5.9. Let a, A ∈ K (K = C, R) such that Re (āA) > 0.
If x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) are sequences of complex numbers
and w̄ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers with the
property that
A Survey on
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S.S. Dragomir
n
X
(5.22)
wi Re [(Ayi − xi ) (x̄i − āȳi )] ≥ 0,
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then one has the inequality
"
(5.23)
n
X
i=1
The constant
1
2
wi |xi |2
n
X
i=1
# 21
wi |yi |2
≤
1
·
2
Pn
i=1
wi Re [Ax̄i yi + āxi ȳi ]
1
[Re (āA)] 2
n
1 |A| + |a| X
≤ ·
w
x
ȳ
i i i .
2 [Re (āA)] 12 i=1
is sharp in the sense that it cannot be replaced by a smaller one.
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Proof. We have, obviously, that
Γ :=
=
n
X
i=1
n
X
wi Re [(Ayi − xi ) (x̄i − āȳi )]
wi Re [Ax̄i yi + āxi ȳi ] −
i=1
n
X
wi |xi |2 − Re (āA)
i=1
n
X
wi |yi |2
i=1
and then, by (5.22), one has
n
X
wi |xi |2 + Re (āA)
i=1
n
X
wi |yi |2 ≤
i=1
n
X
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wi Re [Ax̄i yi + āxi ȳi ]
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i=1
giving
(5.24)
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n
X
1
[Re (āA)]
1
2
1
wi |xi |2 + [Re (āA)] 2
i=1
n
X
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wi |yi |2
i=1
Pn
≤
i=1
wi Re [Ax̄i yi + āxi ȳi ]
1
[Re (āA)] 2
On the other hand, by the elementary inequality
αp2 +
1 2
q ≥ 2pq
α
.
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holding for any p, q ≥ 0 and α > 0, we deduce
(5.25) 2
n
X
i=1
2
wi |xi |
n
X
! 12
2
wi |yi |
i=1
n
X
1
≤
[Re (āA)]
1
2
2
wi |xi | + [Re (āA)]
i=1
1
2
n
X
wi |yi |2 .
i=1
Utilising (5.24) and (5.25), we deduce the first part of (5.23).
The second part is obvious by the fact that for z ∈ C, |Re (z)| ≤ |z| .
Now, assume that the first inequality in (5.23) holds with a constant c > 0,
i.e.,
Pn
n
n
X
X
wi Re [Ax̄i yi + āxi ȳi ]
2
2
(5.26)
wi |xi |
wi |yi | ≤ c · i=1
,
1
2
[Re
(āA)]
i=1
i=1
where a, A, x̄, ȳ satisfy (5.22).
If we choose a = A = 1, y = x 6= 0, then obviously (5.23) holds and from
(5.26) we may get
n
n
X
X
2
wi |xi | ≤ 2c
wi |xi |2 ,
i=1
i=1
giving c ≥ 12 .
The theorem is completely proved.
The following corollary is a natural consequence of the above theorem.
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Corollary 5.10. Let m, M > 0 and x̄, ȳ, w̄ be as in Theorem 5.9 and with the
property that
n
X
(5.27)
wi Re [(M yi − xi ) (x̄i − mȳi )] ≥ 0,
i=1
then one has the inequality
" n
# 21
n
n
X
X
1 M +mX
2
2
(5.28)
≤ · √
wi Re (xi ȳi )
wi |xi |
wi |yi |
2
mM i=1
i=1
i=1
n
X
1 M +m
wi xi ȳi .
≤ · √
2
mM A Survey on
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i=1
The following corollary also holds.
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Corollary 5.11. With the assumptions in Corollary 5.10, then one has the following inequality:
" n
# 12 n
n
X
X
X
2
2
(5.29) 0 ≤
wi |xi |
wi |yi |
−
wi xi ȳi i=1
i=1
i=1
1
" n
#2
n
n
X
X
X
2
2
≤
wi |xi |
wi |yi |
−
wi Re (xi ȳi )
i=1
√
≤
i=1
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i=1
√
√ 2
√ 2
n
M− m X
M− m
√
√
wi Re (xi ȳi ) ≤
2 mM
2
mM
i=1
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n
X
w
x
ȳ
i i i
i=1
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and
(5.30)
2
n
n
X
X
0≤
wi |xi |2
wi |yi |2 − wi xi ȳi i=1
i=1
i=1
"
#2
n
n
n
X
X
X
≤
wi |xi |2
wi |yi |2 −
wi Re (xi ȳi )
n
X
i=1
i=1
2
(M − m)
≤
4mM
5.5.
"
n
X
i=1
#2
wi Re (xi ȳi )
i=1
(M − m)2
≤
4mM
2
n
X
wi xi ȳi .
i=1
A Reverse Inequality for Real Numbers
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
The following result holds [10, Proposition 5.1].
Theorem 5.12. Let a, A ∈ R and x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) be two
sequences with the property that:
(5.31)
ayi ≤ xi ≤ Ayi for each i ∈ {1, . . . , n} .
Then for any w̄ = (w1 , . . . , wn ) a sequence of positive real numbers, one has
the inequality
!2
n
n
n
X
X
X
(5.32)
0≤
wi x2i
wi yi2 −
wi xi yi
i=1
≤
i=1
1
(A − a)2
4
i=1
n
X
i=1
!2
wi yi2
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.
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The constant
1
4
is sharp in (5.32).
Proof. Let us define
I1 :=
A
n
X
wi yi2 −
n
X
i=1
and
I2 :=
!
n
X
wi xi yi
i=1
i=1
n
X
!
wi yi2
i=1
wi xi yi − a
n
X
n
X
!
wi yi2
i=1
(Ayi − xi ) (xi − ayi ) wi .
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
i=1
Then
S.S. Dragomir
I1 = (a + A)
n
X
wi yi2
i=1
n
X
n
X
wi xi yi −
i=1
!2
wi xi yi
n
X
− aA
i=1
!2
wi yi2
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i=1
Contents
and
I2 = (a + A)
n
X
i=1
wi yi2
n
X
wi xi yi −
i=1
n
X
wi x2i
i=1
n
X
wi yi2 − aA
i=1
i=1
giving
(5.33)
n
X
!2
wi yi2
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I1 − I2 =
n
X
i=1
wi x2i
n
X
i=1
wi yi2
−
n
X
Quit
!2
wi yi2
.
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If (5.31) holds, then (Ayi − xi ) (xi − ayi ) ≥ 0 for each i ∈ {1, . . . , n} and thus
I2 ≥ 0 giving
(5.34)
n
X
wi x2i
n
X
i=1
wi yi2
n
X
−
i=1
"
≤
A
n
X
!2
wi yi2
i=1
n
X
wi yi2
−
i=1
!
n
X
wi xi yi
i=1
wi xi yi − a
i=1
n
X
!#
wi yi2
.
i=1
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
If we use the elementary inequality for real numbers u, v ∈ R
uv ≤
(5.35)
1
(u + v)2 ,
4
S.S. Dragomir
then we have for
u := A
Title Page
n
X
wi yi2 −
i=1
n
X
wi xi yi ,
i=1
v :=
n
X
wi xi yi − a
i=1
n
X
Contents
wi yi2
JJ
J
i=1
that
A
n
X
i=1
wi yi2 −
n
X
i=1
!
wi xi yi
n
X
i=1
wi xi yi − a
n
X
Go Back
!
wi yi2
Close
i=1
1
≤ (A − a)2
4
and the inequality (5.32) is proved.
II
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n
X
!2
wi yi2
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Now, assume that (5.32) holds with a constant c > 0, i.e.,
!2
!2
n
n
n
n
X
X
X
X
(5.36)
wi x2i
wi yi2 −
wi xi yi
≤ c (A − a)2
wi yi2 ,
i=1
i=1
i=1
i=1
where a, A, x̄, ȳ satisfy (5.31).
We choose n = 2, w1 = w2 = 1 and let a, A, y1 , y2 , x, α ∈ R such that
ay1 < x1 = Ay1 ,
ay2 = x2 < Ay2 .
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
With these choices, we get from (5.36) that
a2 y12 + a2 y22
2
2
y12 + y22 − A2 y12 + a2 y22 ≤ c (A − a)2 y12 + y22 ,
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which is equivalent to
(A − a)2 y12 y22 ≤ c (A − a)2 y12 + y22
Since we may choose a 6= A, we deduce
.
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2 2
y12 y22 ≤ c y12 + y2
giving, for y1 = y2 = 1, c ≥ 41 .
2
,
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The following corollary is obvious.
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Corollary 5.13. With the above assumptions for a, A, x̄ and ȳ, we have the
inequality
!2
!2
n
n
n
n
X
X
X
X
1
(5.37)
0≤
x2i
yi2 −
yi2 .
xi yi
≤ (A − a)2
4
i=1
i=1
i=1
i=1
Remark 5.3. Condition (5.31) may be replaced by the weaker condition
n
X
(5.38)
wi (Ayi − xi ) (xi − ayi ) ≥ 0
i=1
and the conclusion in Theorem 5.12 will still be valid, i.e., the inequality (5.32)
holds.
For (5.37) to be true it suffices that
n
X
(5.39)
(Ayi − xi ) (xi − ayi ) ≥ 0
i=1
A Reverse Inequality for Complex Numbers
The following result holds [10, Proposition 5.1].
Theorem 5.14. Let a, A ∈ C and x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) ∈ Cn ,
w̄ = (w1 , . . . , wn ) ∈ Rn+ . If
(5.40)
S.S. Dragomir
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J
holds true.
5.6.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
i=1
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wi Re [(Ayi − xi ) (x̄i − āȳi )] ≥ 0,
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then one has the inequality
2
n
n
X
X
2
2
0≤
wi |xi |
wi |yi | − wi xi ȳi i=1
i=1
i=1
!
2
n
X
1
2
2
wi |yi |
.
≤ |A − a|
4
i=1
n
X
(5.41)
The constant
1
4
is sharp in (5.41).
Proof. Consider
"
A1 := Re
A
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
n
X
wi |yi |2 −
i=1
n
X
!
wi xi ȳi
i=1
n
X
wi x̄i yi − ā
i=1
n
X
!#
wi |yi |2
i=1
Contents
and
A2 :=
n
X
"
wi |yi |2 − Re
i=1
n
X
#
wi (Ayi − xi ) (x̄i − āȳi ) .
i=1
Then
A1 =
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n
X
i=1
"
wi |yi |2 − Re A
n
X
wi x̄i yi + ā
i=1
n
X
#
Close
wi xi ȳi
Quit
i=1
2
!2
n
n
X
X
−
wi xi ȳi − Re (āA)
wi |yi |2
i=1
i=1
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and
A2 =
n
X
"
n
X
wi |yi |2 − Re A
i=1
wi x̄i yi + ā
i=1
−
n
X
#
wi xi ȳi
i=1
2
wi |xi |
i=1
n
X
n
X
2
wi |yi | − Re (āA)
i=1
n
X
!2
2
wi |yi |
i=1
giving
(5.42)
A 1 − A2 =
n
X
i=1
2
n
n
X
X
wi |xi |2
wi |yi |2 − wi xi ȳi .
i=1
i=1
If (5.40) holds, then A2 ≥ 0 and thus
n
2
n
n
X
X
X
(5.43)
wi |xi |2
wi |yi |2 − wi xi ȳi i=1
i=1
i=1
"
!
!#
n
n
n
n
X
X
X
X
≤ Re
A
wi |yi |2 −
wi xi ȳi
wi x̄i yi − ā
wi |yi |2
.
i=1
i=1
i=1
Re [z t̄] ≤
then we have for
z := A
n
X
i=1
1
|z − t|2 ,
4
2
wi |yi | −
n
X
i=1
S.S. Dragomir
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i=1
If we use the elementary inequality for complex numbers z, t ∈ C
(5.44)
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Type Discrete Inequalities
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wi xi ȳi ,
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t :=
n
X
wi xi ȳi − a
n
X
i=1
wi |yi |2
i=1
that
"
(5.45)
Re
A
n
X
2
wi |yi | −
i=1
n
X
!
wi xi ȳi
i=1
n
X
wi x̄i yi − ā
n
X
i=1
1
≤ |A − a|2
4
!#
2
wi |yi |
i=1
n
X
!2
wi |yi |2
i=1
and the inequality (5.41) is proved.
Now, assume that (5.41) holds with a constant c > 0, i.e.,
2
!2
n
n
n
n
X
X
X
X
(5.46)
wi |xi |2
wi |yi |2 − wi xi ȳi ≤ c |A − a|2
wi |yi |2 ,
i=1
i=1
i=1
i=1
where x̄, ȳ, a, A satisfy (5.40).
P
P
Consider
ȳ ∈ Cn , ni=1 |yi |2 wi = 1, a 6= A, m̄ ∈ Cn , ni=1 wi |mi |2 = 1
Pn
with i=1 wi yi mi = 0. Define
A+a
A+a
xi :=
yi +
mi ,
2
2
i ∈ {1, . . . , n} .
Then
n
X
i=1
n
A − a 2 X
wi (Ayi − xi ) (x̄i − āyi ) = wi (yi − mi ) (ȳi − m̄i ) = 0
2 i=1
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
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and thus the condition (5.40) is fulfilled.
From (5.46) we deduce
2
2
n n X
X
A + a
A
−
a
A
+
a
A
−
a
y
+
m
w
−
y
+
m
ȳ
w
i
i
i
i
i
i i
2
2
2
2
i=1
i=1
≤ c |A − a|2
and since
n
X
i=1
and
2 A + a 2 A − a 2
A+a
A
−
a
−
yi +
mi = wi 2 2
2
2 S.S. Dragomir
n 2 X A + a
A + a 2
A−a
yi +
mi ȳi wi = 2
2
2
i=1
then by (5.46) we get
giving c ≥
1
4
|A − a|2
≤ c |A − a|2
4
and the theorem is completely proved.
The following corollary holds.
Corollary 5.15. Let a, A ∈ C and x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) ∈ Cn
be with the property that
(5.47)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
i=1
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Re [(Ayi − xi ) (x̄i − āȳi )] ≥ 0,
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then one has the inequality
(5.48)
0≤
n
X
i=1
The constant
1
4
2
n
n
X
X
1
|xi |2
|yi |2 − xi ȳi ≤ |A − a|2
4
i=1
i=1
n
X
!2
|yi |2
.
i=1
is best in (5.48).
Remark 5.4. A sufficient condition for both (5.40) and (5.47) to hold is
Re [(Ayi − xi ) (x̄i − āȳi )] ≥ 0
(5.49)
for any i ∈ {1, . . . , n} .
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
5.7.
Shisha-Mond Type Inequalities
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As some particular case for bounds on differences of means, O. Shisha and B.
Mond obtained in 1967 (see [23]) the following reverse of (CBS) − inequality:
Theorem 5.16. Assume that ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) are such
that there exists a, A, b, B > 0 with the property that:
(5.50)
a ≤ aj ≤ A and b ≤ bj ≤ B for any j ∈ {1, . . . , n}
(5.51)
j=1
a2j
n
X
j=1
b2j
−
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then we have the inequality
n
X
Contents
n
X
j=1
!2
aj b j
r
≤
A
−
b
r !2 n
n
X
X
a
aj b j
b2j .
B
j=1
j=1
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The equality holds in (5.51) if and only if there exists a subsequence (k1 , . . . , kp )
of (1, 2, . . . , n) such that
12 32
A
n
B
=1+
,
p
a
b
akµ = A, bkµ = b (µ = 1, . . . , p) and ak = a, bk = B for every k distinct from
all kµ .
Using another result stated for weighted means in [23], we may prove the
following reverse of the (CBS) −inequality.
Theorem 5.17. Assume that ā, b̄ are positive sequences and there exists γ, Γ >
0 with the property that
ai
(5.52)
0<γ≤
≤ Γ < ∞ for any i ∈ {1, . . . , n} .
bi
0≤
(5.53)
i=1
a2i
n
X
! 21
b2i
i=1
−
n
X
i=1
n
(Γ − γ)2 X 2
ai b i ≤
b.
4 (γ + Γ) j=1 j
The equality holds in (5.53) if and only if there exists a subsequence (k1 , . . . , kp )
of (1, 2, . . . , n) such that
p
X
m=1
n
b2km
S.S. Dragomir
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Then we have the inequality
n
X
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
Γ + 3γ X 2 akm
ak
=
bj ,
= Γ (m = 1, . . . , p) and
=γ
4 (γ + Γ) j=1
b km
bk
for every k distinct from all km .
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Proof. In [23, p. 301], Shisha and Mond have proved the following weighted
inequality
! 12
n
n
X
X
(C − c)2
2
qj x j
−
qj x j ≤
,
(5.54)
0≤
4
(c
+
C)
j=1
j=1
P
provided qj ≥ 0 (j = 1, . . . , n) with nj=1 qj = 1 and 0 < c ≤ xj < C < ∞
for any j ∈ {1, . . . , n} .
Equality holds in (5.54) if and only if there exists a subsequence (k1 , . . . , kp )
of (1, 2, . . . , n) such that
(5.55)
p
X
m=1
q km =
C + 3c
,
4 (c + C)
xkm = C (m = 1, 2, . . . , p) and xk = c for every k distinct from all km .
If in (5.54) we choose
b2j
aj
xj = , qj = Pn 2 , j ∈ {1, . . . , n} ;
bj
k=1 bk
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S.S. Dragomir
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then we get
! 12 Pn
Pn
2
a
(Γ − γ)2
j
j=1 aj bj
P
Pj=1
−
≤
,
n
n
2
2
4 (γ + Γ)
k=1 bk
k=1 bk
giving the desired inequality (5.53).
The case of equality follows by the similar case in (5.54) and we omit the
details.
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5.8.
Zagier Type Inequalities
The following result was obtained by D. Zagier in 1995, [24].
Lemma 5.18. Let f, g : [0, ∞) → R be monotone decreasing nonnegative
functions on [0, ∞). Then
R∞
R∞
Z ∞
f
(x)
F
(x)
dx
g (x) G (x) dx
0R
R ∞
,
(5.56)
f (x) g (x) dx ≥ 0
∞
max 0 F (x) dx, 0 G (x) dx
0
A Survey on
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Type Discrete Inequalities
for any integrable functions F, G : [0, ∞) → [0, 1] .
Proof. We will follow the proof in [24].
For all x ≥ 0 we have
Z ∞
Z ∞
Z ∞
f (t) F (t) dt = f (x)
F (t) dt +
[f (t) − f (x)] F (t) dt
0
0
0
Z ∞
Z x
≤ f (x)
F (t) dt +
[f (t) − f (x)] dt
0
0
Rx
R∞
and hence, since 0 G (t) dt is bounded from above by both x and 0 G (t) dt,
Z ∞
Z x
f (t) F (t) dt ·
G (t) dt
0
0
Z ∞
Z ∞
Z x
≤ xf (x)
F (t) dt +
G (t) dt ·
[f (t) − f (x)] dt
0
0
0
Z ∞
Z x
Z ∞
≤ max
F (t) dt,
G (t) dt ·
f (t) dt.
0
0
0
S.S. Dragomir
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Now, we multiply
by −dg (x) Rand integrate by parts from 0 to ∞. The left hand
R∞
∞
side gives −∞ f (t) F (t) dt · −∞ g (t) G (t) dt, the right hand side gives
Z ∞
Z ∞
Z ∞
max
F (t) dt,
G (t) dt ·
f (t) g (t) dt,
0
0
0
and the inequality remains true because the measure −dg (x) is nonnegative.
The following particular case is a reverse of the (CBS) −integral inequality
obtained by D. Zagier in 1977, [25].
Corollary 5.19. If f, g : [0, ∞) → [0, ∞) are decreasing function on [0, ∞),
then
Z ∞
Z ∞
Z ∞
(5.57) max f (0)
g (t) dt, g (0)
f (t) dt ·
f (t) g (t) dt
0
0
0
Z ∞
Z ∞
2
≥
f (t) dt
g 2 (t) dt.
0
0
Remark 5.5. The following weighted version of (5.56) may be proved in a similar way, as noted by D. Zagier in [25]
Z ∞
(5.58)
w (t) f (t) g (t) dt
0
R∞
R∞
w (t) f (t) F (t) dt 0 w (t) f (t) G (t) dt
0
R ∞
,
R∞
≥
max 0 w (t) F (t) dt, 0 w (t) G (t) dt
provided w (t) > 0 on [0, ∞), f, g : [0, ∞) → [0, ∞) are monotonic decreasing
and F, G : [0, ∞) → [0, 1] are integrable on [0, ∞).
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We may state and prove the following discrete inequality.
Theorem 5.20. Consider the sequences of real numbers ā = (a1 , . . . , an ), b̄ =
(b1 , . . . , bn ) , p̄ = (p1 , . . . , pn ), q̄ = (q1 , . . . , qn ) and w̄ = (w1 , . . . , wn ) .
If
(i) ā and b̄ are decreasing and nonnegative;
(ii) pi , qi ∈ [0, 1] and wi ≥ 0 for any i ∈ {1, . . . , n} ,
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
then we have the inequality
(5.59)
n
X
i=1
Pn
wi ai bi ≥
P
wi pi ai ni=1 wi qi bi
i=1
P
P
.
max { ni=1 wi pi , ni=1 wi qi }
Proof. Consider the functions f, g, F, G, W : [0, ∞) → R given by


b1 , t ∈ [0, 1)

a
,
t
∈
[0,
1)


1










 a2 , t ∈ [1, 2)
 b2 , t ∈ [1, 2)




..
.
f (t) =
,
g (t) =
,
.
.
.








an , t ∈ [n − 1, n)


bn , t ∈ [n − 1, n)









0
t ∈ [n, ∞)
0
t ∈ [n, ∞)
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
p1 ,






p2 ,




..
F (t) =
.





pn ,





0
and
t ∈ [0, 1)
t ∈ [1, 2)
,
t ∈ [n − 1, n)
t ∈ [n, ∞)

w1 ,






w2 ,




..
W (t) =
.





wn ,





0

q1 ,






q2 ,




..
G (t) =
.





qn ,





0
t ∈ [0, 1)
t ∈ [1, 2)
,
t ∈ [n − 1, n)
t ∈ [n, ∞)
t ∈ [0, 1)
A Survey on
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t ∈ [1, 2)
S.S. Dragomir
.
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t ∈ [n − 1, n)
Contents
t ∈ [n, ∞)
We observe that, the above functions satisfy the hypothesis of Remark 5.5 and
since, for example,
Z ∞
Z ∞
n Z i
X
w (t) f (t) g (t) dt =
w (t) f (t) g (t) dt +
w (t) f (t) g (t) dt
0
=
i=1
n
X
i−1
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wk ak bk ,
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k=1
then by (5.58) we deduce the desired inequality (5.59).
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Remark 5.6. A similar inequality for sequences under some monotonicity assumptions for p̄ and q̄ was obtained in 1995 by J. Pečarić in [26].
The following reverse of the (CBS) −discrete inequality holds.
Theorem 5.21. Assume that ā, b̄ are decreasing nonnegative sequences with
a1 , b1 6= 0 and w̄ a nonnegative sequence. Then
( n
) n
n
n
n
X
X
X
X
X
(5.60)
wi a2i
wi b2i ≤ max b1
wi ai , a1
wi bi
wi ai bi .
i=1
i=1
i=1
i=1
The proof follows by Theorem 5.20 on choosing pi =
[0, 1] , i ∈ {1, . . . , n} . We omit the details.
i=1
ai
a1
∈ [0, 1] , qi =
bi
b1
∈
Remark 5.7. When wi = 1, we recapture Alzer’s result from 1992, [27].
5.9.
A Reverse Inequality in Terms of the sup −Norm
Lemma 5.22. Let ᾱ = (α1 , . . . , αn ) and x̄ = (x1 , . . . , xn ) be sequences of
complex numbers
and p̄ = (p1 , . . . , pn ) a sequence of nonnegative real numbers
Pn
such that i=1 pi = 1. Then one has the inequality
n
n
n
X
X
X
(5.61) pi αi xi −
pi αi
pi xi i=1
i=1
i=1

!2 
n
n
X
X
≤ max |∆αi | max |∆xi | 
i2 p i −
ipi  ,
i=1,n−1
S.S. Dragomir
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The following result has been proved in [11].
i=1,n−1
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where ∆αi is the forward difference, i.e., ∆αi := αi+1 − αi .
Inequality (5.61) is sharp in the sense that the constant C = 1 in the right
membership cannot be replaced be a smaller one.
Proof. We shall follow the proof in [11]. We start with the following identity
n
X
i=1
pi αi xi −
n
X
pi αi
i=1
n
X
i=1
n
1X
pi pj (αi − αj ) (xi − xj )
2 i,j=1
X
=
pi pj (αi − αj ) (xi − xj ) .
pi xi =
1≤i<j≤n
As i < j, we can write that
A Survey on
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S.S. Dragomir
αj − αi =
j−1
X
∆αk
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k=i
and
xj − xi =
j−1
X
Contents
JJ
J
∆xk .
k=i
Using the generalised triangle inequality, we have successively
j−1
j−1
n
n
n
X
X
X
X
X
X
pi αi xi −
pi αi
pi xi = pi pj
∆αk
∆xk i=1
i=1
i=1
1≤i<j≤n
k=i
k=i
j−1
j−1
X
X
X
≤
pi pj ∆αk ∆xk 1≤i<j≤n
k=i
k=i
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X
≤
pi pj
j−1
X
1≤i<j≤n
|∆αk |
k=i
j−1
X
|∆xk | := A.
k=i
Note that
|∆αk | ≤ max |∆αs |
1≤s≤n−1
and
|∆xk | ≤ max |∆xs |
1≤s≤n−1
A Survey on
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for all k = i, . . . , j − 1 and then by summation
j−1
X
S.S. Dragomir
|∆αk | ≤ (j − i) max |∆αs |
1≤s≤n−1
k=i
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and
j−1
X
Contents
|∆xk | ≤ (j − i) max |∆xs | .
1≤s≤n−1
k=i
Taking into account the above estimations, we can write
"
#
X
A≤
pi pj (j − i)2 max |∆αs | max |∆xs | .
1≤s≤n−1
1≤i<j≤n
1≤s≤n−1
1≤i<j≤n
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As a simple calculation shows that
X
JJ
J
2
pi pj (j − i) =
n
X
i=1
2
i pi −
n
X
i=1
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!2
ipi
,
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inequality (5.61) is proved.
To prove the sharpness of the constant, let us assume that (5.61) holds with
a constant C > 0, i.e.,
n
n
n
X
X
X
(5.62) pi αi xi −
pi αi
pi xi i=1
i=1
i=1

!2 
n
n
X
X
i2 p i −
ipi  .
≤ C max |∆αi | max |∆xi | 
i=1,n−1
i=1,n−1
i=1
i=1
Now, choose the sequences αk = α + kβ (β 6= 0) and xk = x + ky (y 6= 0) ,
k ∈ {1, . . . , n} to get
n
n
n
n
X
1 X
X
X
p
α
x
−
p
α
p
x
=
p
p
(i
−
j)
βy
i i i
i i
i i
i j
2
i=1
i=1
i=1
i,j=1

!2 
n
n
X
X
i2 p i −
ipi 
= |β| |y| 
i=1
i=1
and

n
X

max |∆αi | max |∆xi |
i2 p i −
i=1,n−1
i=1,n−1
i=1
n
X
S.S. Dragomir
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!2 
ipi
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i=1

n
X

= |β| |y|
i2 p i −
i=1
n
X
i=1
!2 
ipi
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
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and then, by (5.62), we get C ≥ 1.
The following reverse of the (CBS) −inequality holds [12].
Theorem 5.23. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with ai 6= 0, (i = 1, . . . , n) . Then one has the inequality
0≤
n
X
i=1
a2i
n
X
i=1
b2i −
n
X
!2
ai bi
i=1

2 X
n
n
X
bk
2

i2 a2i −
ai
≤ max ∆
ak
k=1,n−1
i=1
i=1
n
X
!2 
ia2i
.
i=1
The constant C = 1 is sharp in the sense that it cannot be replaced by a smaller
constant.
Proof. Follows by Lemma 5.22 on choosing
a2i
bi
bi
pi = Pn
, αi = , xi = , i ∈ {1, . . . , n}
2
ai
ai
k=1 ak
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and performing some elementary calculations.
We omit the details.
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5.10.
A Reverse Inequality in Terms of the 1−Norm
The following result has been obtained in [13].
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Lemma 5.24. Let ᾱ = (α1 , . . . , αn ) and x̄ = (x1 , . . . , xn ) be sequences of
complex numbers
and p̄ = (p1 , . . . , pn ) a sequence of nonnegative real numbers
P
such that ni=1 pi = 1. Then one has the inequality
n
n
n
X
X
X
(5.63) pi αi xi −
pi αi
pi xi i=1
i=1
i=1
n
n−1
n−1
X
X
1X
≤
pi (1 − pi )
|∆αi |
|∆xi | ,
2 i=1
i=1
i=1
where ∆αi := αi+1 − αi is the forward difference.
The constant 12 is sharp in the sense that it cannot be replaced by a smaller
constant.
i=1
≤
pi pj
1≤i<j≤n
j−1
X
|∆αk |
k=i
j−1
X
l=i
It is obvious that for all 1 ≤ i < j ≤ n − 1, we have that
j−1
X
k=i
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i=1
X
|∆αk | ≤
n−1
X
k=1
S.S. Dragomir
Title Page
Proof. We shall follow the proof in [13].
As in the proof of Lemma 5.22 in Section 5.9, we have
n
n
n
X
X
X
(5.64)
pi αi xi −
pi αi
pi xi i=1
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|∆xl | := A.
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|∆αk |
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and
j−1
X
|∆xl | ≤
n−1
X
|∆xl | .
l=1
l=i
Utilising these and the definition of A, we conclude that
A≤
(5.65)
n−1
X
|∆αk |
k=1
n−1
X
l=1
X
|∆xl |
pi pj .
1≤i<j≤n
A Survey on
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Now, let us observe that
" n
#
X
1 X
pi pj −
pi pj
pi pj =
2
i,j=1
i=j
1≤i<j≤n
" n
#
n
n
X
1 X X
=
pi
pj −
p2i
2 i=1 j=1
i=1
S.S. Dragomir
X
(5.66)
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n
=
1X
pi (1 − pi ) .
2 i=1
Making use of (5.64) – (5.66), we deduce the desired inequality (5.63).
To prove the sharpness of the constant 12 , let us assume that (5.63) holds with
a constant C > 0. That is
n
n
n
n
n−1
n−1
X
X
X
X
X
X
(5.67) pi αi xi −
pi αi
pi xi ≤ C
pi (1 − pi )
|∆αi |
|∆xi |
i=1
i=1
i=1
i=1
i=1
i=1
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for all αi , xi , pi (i = 1, . . . , n) as above and n ≥ 1.
Choose in (5.63) n = 2 and compute
2
X
pi αi xi −
i=1
2
X
pi αi
i=1
2
X
i=1
2
1X
pi xi =
pi pj (αi − αj ) (xi − xj )
2 i,j=1
X
=
pi pj (αi − αj ) (xi − xj )
1≤i<j≤2
= p1 p2 (α1 − α2 ) (x1 − x2 ) .
Also
2
X
i=1
pi (1 − pi )
2
X
i=1
|∆αi |
2
X
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S.S. Dragomir
|∆xi | = (p1 p2 + p1 p2 ) |α1 − α2 | |x1 − x2 | .
i=1
Substituting in (5.67), we obtain
p1 p2 |α1 − α2 | |x1 − x2 | ≤ 2Cp1 p2 |α1 − α2 | |x1 − x2 | .
If we assume that p1 , p2 > 0, α1 6= α2 , x1 6= x2 , then we obtain C ≥ 21 , which
proves the sharpness of the constant 21 .
We are now able to state the following reverse of the (CBS) −inequality
[12].
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Theorem 5.25. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with ai 6= 0 (i = 1, . . . , n) . Then one has the inequality
!2
n
n
n
X
X
X
(5.68)
0≤
a2i
b2i −
ai b i
i=1
i=1
i=1
" n−1 #2
X
X
∆ bk ≤
a2i a2j .
ak
1≤i<j≤n
k=1
The constant C = 1 is sharp in (5.68), in the sense that it cannot be replaced
by a smaller constant.
A Survey on
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S.S. Dragomir
Proof. We choose
a2
pi = Pn i
2
k=1 ak
,
αi = xi =
bi
, i ∈ {1, . . . , n}
ai
in (5.63) to get
P
Pn 2
2
bi
( ni=1 ai bi )
i=1
0 ≤ Pn
−
P
2
2
( nk=1 a2k )
k=1 ak
Pn 2 a2
Pn i 2
a
1
−
i=1 i
1
k=1 ak
Pn
≤ ·
2
2
k=1 ak
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!2
b
j
∆
a
j
j=1
Pn 2 Pn
!2
n−1 2
2
X
a
(
a
−
a
)
1
b
j
i
∆
= · i=1 iPn k=12 2k
2
a
( k=1 ak )
j
j=1
n−1 X
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which is clearly equivalent to
0≤
n
X
a2i
i=1

1
≤ 
2
Since

1
2
n
X
n
X
b2i −
i=1
n
X
ai bi
i=1
!2
a2k
−
n
X
i=1
k=1
n
X
!2
!2
a2k
−
k=1
n
X

n X
bj 4
ai
∆ aj !2
.
j=1

a4i 
X
=
i=1
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a2i a2j
1≤i<j≤n
S.S. Dragomir
the inequality (5.68) is thus proved.
5.11.
A Reverse Inequality in Terms of the p−Norm
The following result has been obtained in [14].
Lemma 5.26. Let ᾱ = (α1 , . . . , αn ) and x̄ = (x1 , . . . , xn ) be sequences of
complex numbers
and p̄ = (p1 , . . . , pn ) a sequence of nonnegative real numbers
P
such that ni=1 pi = 1. Then one has the inequality
n
n
n
X
X
X
(5.69) pi αi xi −
pi αi
pi xi i=1
i=1
i=1
! p1 n−1
! 1q
n−1
X
X
X
≤
(i − j) pi pj
|∆αk |p
|∆xk |q
,
1≤j<i≤n
k=1
k=1
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where p > 1, p1 + 1q = 1.
The constant C = 1 in the right hand side of (5.69) is sharp in the sense that
it cannot be replaced by a smaller constant.
Proof. We shall follow the proof in [14].
As in the proof of Lemma 5.22 in Section 5.9, we have
n
n
n
X
X
X
(5.70)
pi αi xi −
pi αi
pi xi i=1
i=1
i=1
≤
X
i−1
X
pi pj
1≤j<i≤n
i−1
X
|∆αk |
k=j
|∆xl | := A.
A Survey on
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S.S. Dragomir
l=j
Using Hölder’s discrete inequality, we can state that
i−1
X
1
|∆αk | ≤ (i − j) q
k=j
i−1
X
where p > 1,
(5.71)
+
A≤
1
q
JJ
J
k=j
|∆xl | ≤ (i − j)
1
p
l=j
1
p
Contents
|∆αk |p
and
i−1
X
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! p1
i−1
X
! 1q
|∆xl |q
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l=j
= 1, and then we get
X
1≤j<i≤n
pi pj (i − j)
i−1
X
k=j
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! p1
p
|∆αk |
i−1
X
k=j
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! 1q
q
|∆xk |
.
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Since
i−1
X
p
|∆αk | ≤
n−1
X
k=j
and
i−1
X
|∆αk |p
k=1
q
|∆xk | ≤
k=j
n−1
X
|∆xk |q ,
k=1
for all 1 ≤ j < i ≤ n, then by (5.70) and (5.71) we deduce the desired inequality (5.69).
To prove the sharpness of the constant, let us assume that (5.69) holds with
a constant C > 0. That is,
n
n
n
X
X
X
(5.72) pi αi xi −
pi αi
pi xi i=1
i=1
i=1
! p1 n−1
! 1q
n−1
X
X
X
|∆xk |q
.
≤C
(i − j) pi pj
|∆αk |p
1≤j<i≤n
k=1
k=1
Note that, for n = 2, we have
2
2
2
X
X
X
p
α
x
−
p
α
p
x
i i i
i i
i i = p1 p2 |α1 − α2 | |x1 − x2 |
i=1
i=1
1≤j<i≤2
S.S. Dragomir
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and
X
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(i − j) pi pj
1
X
k=1
! p1
p
|∆αk |
1
X
k=1
Page 182 of 288
! 1q
q
|∆xk |
= p1 p2 |α1 − α2 | |x1 − x2 | .
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Therefore, from (5.72), we obtain
p1 p2 |α1 − α2 | |x1 − x2 | ≤ Cp1 p2 |α1 − α2 | |x1 − x2 |
for all α1 6= α2 , x1 6= x2 , p1 p2 > 0, giving C ≥ 1.
We are able now to state the following reverse of the (CBS) −inequality.
Theorem 5.27. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with ai 6= 0, (i = 1, . . . , n) . Then one has the inequality
!2
n
n
n
X
X
X
0≤
a2i
b2i −
ai b i
i=1
≤
i=1
i=1
p ! p1
n−1 X
b
∆ k ak k=1
1
p
q ! 1q
n−1 X
b
∆ k ak k=1
X
(i − j) a2i a2j ,
1≤j<i≤n
1
q
where p > 1, + = 1.
The constant C = 1 is sharp in the above sense.
Proof. Follows by Lemma 5.26 for
a2i
bi
pi = Pn
, αi = xi = , i ∈ {1, . . . , n} .
2
ai
k=1 ak
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The following corollary is a natural consequence of Theorem 5.27 for p =
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Corollary 5.28. With the assumptions of Theorem 5.27 for ā and b̄, we have
!2
n
n
n
X
X
X
0≤
a2i
b2i −
ai b i
i=1
i=1
i=1
2 X
n−1 X
b
k
∆
≤
(i − j) a2i a2j .
ak 1≤j<i≤n
k=1
5.12.
A Reverse Inequality Via an Andrica-Badea Result
The following result is due to Andrica and Badea [15, p. 16].
Lemma 5.29. Let x̄ = (x1 , . . . , xn ) ∈ I n = [m, M ]n be a sequence of real
numbers and let S be the subset of {1, . . . , n} that minimises the expression
X
1 (5.73)
p i − Pn ,
2 i∈S
Pn
where Pn :=
numbers. Then

(5.74)
maxn 
x̄∈I
i=1
pi > 0, p̄ = (p1 , . . . , pn ) is a sequence of nonnegative real
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n
1 X 2
pi xi −
Pn i=1
!2 
n
1 X
pi xi 
Pn i=1
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!
=
X
i∈S
p i Pn −
X
i∈S
pi
(M − m)2
.
Pn2
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Proof. We shall follow the proof in [15, p. 161].
Define
n
n
1 X 2
1 X
Dn (x̄, p̄) :=
pi xi −
pi xi
Pn i=1
Pn i=1
1 X
pi pj (xi − xj )2 .
=
Pn 1≤i<j≤n
!2
Keeping in mind the convexity of the quadratic function, we have
Dn (αx̄ + (1 − α) ȳ, p̄)
1 X
= 2
pi pj [αxi + (1 − α) yi − αxj − (1 − α) yj ]2
Pn 1≤i<j≤n
1 X
= 2
pi pj [α (xi − xj ) + (1 − α) (yi − yj )]2
Pn 1≤i<j≤n
1 X
≤ 2
pi pj α (xi − xj )2 + (1 − α) (yi − yj )2
Pn 1≤i<j≤n
= αDn (x̄, p̄) + (1 − α) Dn (ȳ, p̄) ,
hence Dn (·, p̄) is a convex function on I n .
Using a well known theorem (see for instance [16, p. 124]), we get that the
maximum ofDn (·, p̄) is attained on the boundary of I n .
Let S, S̄ be the partition of {1, . . . , n} such that the maximum of Dn (·, p̄)
is obtained for x̄0 = (x01 , . . . , x0n ) , where x0i = m if i ∈ S̄ and x0i = M if i ∈ S.
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In this case we have
(5.75)
Dn (x̄0 , p̄) =
1
Pn2
X
pi pj (xi − xj )2
1≤i<j≤n
!
X
(M − m)2 X
=
p i Pn −
pi .
Pn2
i∈S
i∈S
The expression
!
X
p i Pn −
i∈S
X
pi
i∈S
is a maximum when the set S minimises the expression
X
1 p i − Pn .
2 i∈S
From (5.75) it follows that Dn (x̄, p̄) is also a maximum and the proof of the
above lemma is complete.
The following reverse result of the (CBS) −inequality holds.
Theorem 5.30. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with ai 6= 0 (i = 1, . . . , n) and
(5.76)
−∞ < m ≤
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bi
≤ M < ∞ for each i ∈ {1, . . . , n} .
ai
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Let S be the subset of {1, . . . , n} that minimizes the expression
n
X
X
1
2
2
(5.77)
ai −
ai ,
2
i=1
i∈S
and denote S̄ := {1, . . . , n} \ S. Then we have the inequality
!2
n
n
n
X
X
X
(5.78)
0≤
a2i
b2i −
ai b i
i=1
i=1
2
≤ (M − m)
X
a2i
i∈S
≤
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i=1
X
S.S. Dragomir
i∈S̄
n
X
1
(M − m)2
4
a2i
!2
a2i
.
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i=1
Proof. The proof of the second inequality in (5.78) follows by Lemma 5.29 on
choosing pi = a2i , xi = abii , i ∈ {1, . . . , n} .
The third inequality is obvious as
!
n
X X
X
X
X
a2i
a2i =
a2i
a2j −
a2i
i∈S
i∈S̄
j=1
i∈S
1
≤
4
X
i∈S
a2i
+
i∈S
n
X
j=1
!2
a2j
−
X
i∈S
a2i
1
=
4
n
X
!2
a2j
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5.13.
A Refinement of Cassels’ Inequality
In 1914, P. Schweitzer [18] proved the following result.
Theorem 5.31. If ā = (a1 , . . . , an ) is a sequence of real numbers such that 0 <
m ≤ ai ≤ M < ∞ (i ∈ {1, . . . , n}) , then
!
!
n
n
(M + m)2
1X
1X 1
≤
(5.79)
ai
.
n i=1
n i=1 ai
4mM
In 1972, A. Lupaş [17] proved the following refinement of Schweitzer’s result which gives the best bound for n odd as well.
Theorem 5.32. With the assumptions in Theorem 5.31, one has
n
n+1 n
n
X
X
M + n+1
m
M + n2 m
1
2
2
2
,
(5.80)
ai
≤
M
m
a
i
i=1
i=1
where [·] is the integer part.
In 1988, Andrica and Badea [15] established a weighted version of Schweitzer
and Lupaş inequalities via the use of the following weighted version of the
Grüss inequality [15, Theorem 2].
Theorem 5.33. If m1 ≤ ai ≤ M1 , m2 ≤ bi ≤ M2 (i ∈ {1, . . . , n}) and S is the
subset of {1, . . . , n} which minimises the expression
X
1
(5.81)
p
−
P
i
n ,
2 i∈S
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P
where Pn := ni=1 pi > 0, then
n
n
n
X
X
X
(5.82)
p i ai b i −
p i ai ·
p i bi Pn
i=1
i=1
i=1
!
≤ (M1 − m1 ) (M2 − m2 )
X
p i Pn −
i∈S
X
pi .
i∈S
1
≤ Pn2 (M1 − m1 ) (M2 − m2 ) .
4
Proof. Using the result in Lemma 5.29, Section 5.12, we have
!2
!
n
n
X
1 X 2
1 X
(M1 − m1 )2 X
pi
(5.83)
p i ai −
p i ai
≤
p i Pn −
Pn i=1
Pn i=1
Pn2
i∈S
i∈S
and
(5.84)
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1
Pn
n
X
i=1
pi b2i −
1
Pn
n
X
i=1
!2
p i bi
2
≤
X
(M2 − m2 ) X
p
P
−
pi
i
n
Pn2
i∈S
i∈S
and since
!
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(5.85)
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n
n
n
1 X
1 X
1 X
p i ai b i −
p i ai ·
p i bi
Pn i=1
Pn i=1
Pn i=1

!2 
n
n
X
X
1
1
≤
pi a2i −
p i ai 
Pn i=1
Pn i=1
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
n
1 X 2

×
p i bi −
Pn i=1
n
1 X
pi bi
Pn i=1
!2 
,
the first part of (5.82) holds true.
The second part follows by the elementary inequality
1
(a + b)2 , a, b ∈ R
4
P
P
for the choices a := i∈S pi , b := Pn − i∈S pi .
ab ≤
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We are now able to state and prove the result of Andrica and Badea [15,
Theorem 4], which is related to Schweitzer’s inequality.
S.S. Dragomir
Theorem 5.34. If 0 < m ≤ ai ≤ M < ∞, i ∈ {1, . . . , n} and S is a subset of
{1, . . . , n} that minimises the expression
X
Pn pi −
,
2 Title Page
i∈S
then we have the inequality
! n
!
!
n
2 X
X
X pi
X
(M
−
m)
(5.86)
p i ai
≤ Pn2 +
p i Pn −
pi
a
Mm
i
i=1
i=1
i∈S
i∈S
(M + m)2 2
Pn .
≤
4M m
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Proof. We shall follow the proof in [15]. We obtain from Theorem 5.32 with
bi = a1i , m1 = m, M1 = m, m2 = M1 , M2 = m1 , the following estimate
!
n
n
X
X
pi 1
1 X
2 X
p i ai
−
p i Pn −
pi ,
Pn −
≤ (M − m)
ai m M
i=1
i=1
i∈S
i∈S
that leads, in a simple manner, to (5.86).
We may now prove the following reverse result for the weighted (CBS) −
inequality that improves the additive version of Cassels’ inequality.
Theorem 5.35. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be two sequences of positive real numbers with the property that
(5.87)
bi
0<m≤
≤ M < ∞ for each i ∈ {1, . . . , n} ,
ai
(5.89)
i=1
i=1
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then one has the inequality
pi a2i
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i∈S
n
X
S.S. Dragomir
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and p̄ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that Pn :=
P
n
i=1 pi > 0. If S is a subset of {1, . . . , n} that minimises the expression
n
X
X
1
(5.88)
p
a
b
−
p
a
b
i i i
i i i
2
n
X
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pi b2i
−
n
X
Close
!2
p i ai b i
Quit
i=1
n
X
X
(M − m)2 X
p i ai b i
p i ai b i −
p i ai b i
≤
Mm
i=1
i∈S
i∈S
!
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(M − m)2
≤
4M m
n
X
!2
.
p i ai b i
i=1
Proof. Applying Theorem 5.34 for ai = xi , pi = qi xi we may deduce the
inequality
(5.90)
n
X
i=1
qi x2i
n
X
i=1
qi −
n
X
!2
qi x i
i=1
!
≤
n
X
X
(M − m)2 X
qi x i
qi x i −
qi x i ,
Mm
i=1
i∈S
i∈S
P
provided qi ≥ 0, ni=1 qi > 0, 0 < m ≤ xi ≤ M < ∞, for i ∈ {1, . . . , n} and
S is a subset of {1, . . . , n} that minimises the expression
n
X
X
1
(5.91)
qi x i −
qi x i .
2
i∈S
i=1
Now, if in (5.90) we choose qi = pi a2i , xi =
we deduce the desired result (5.89).
bi
ai
∈ [m, M ] for i ∈ {1, . . . , n} ,
The following corollary provides a refinement of Cassels’ inequality.
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Corollary 5.36. With the assumptions of Theorem 5.35, we have the inequality
Pn
Pn
2
2
i=1 pi ai
i=1 pi bi
1≤
(5.92)
P
2
( ni=1 pi ai bi )
P
P
(M − m)2
i∈S pi ai bi
i∈S pi ai bi
≤1+
1 − Pn
· Pn
Mm
i=1 pi ai bi
i=1 pi ai bi
≤
(M + m)2
.
4M m
The case of the “unweighted” Cassels’ inequality is embodied in the following corollary as well.
Corollary 5.37. Assume that ā and b̄ satisfy (5.88). If S is a subset of {1, . . . , n}
that minimises the expression
n
X
1X
(5.93)
ai b i −
ai b i 2
i∈S
i=1
then one has the inequality
Pn 2 Pn 2
ai i=1 bi
(5.94)
1 ≤ i=1
Pn
2
( i=1 ai bi )
P
P
(M − m)2
i∈S ai bi
i∈S ai bi
· Pn
1 − Pn
≤1+
Mm
i=1 ai bi
i=1 ai bi
(M + m)2
≤
.
4M m
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In particular, we may obtain the following refinement of the Pólya-Szegö’s
inequality.
Corollary 5.38. Assume that
(5.95)
0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for i ∈ {1, . . . , n} .
If S is a subset of {1, . . . , n} that minimises the expression (5.93), then one has
the inequality
Pn 2 Pn 2
ai i=1 bi
(5.96)
1 ≤ i=1
Pn
2
( i=1 ai bi )
P
P
(AB − ab)2
i∈S ai bi
i∈S ai bi
≤1+
· Pn
1 − Pn
abAB
i=1 ai bi
i=1 ai bi
(AB + ab)2
≤
.
4abAB
5.14.
Two Reverse Results Via Diaz-Metcalf Results
In [19], J.B. Diaz and F.T. Metcalf proved the following inequality for sequences
of complex numbers.
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Lemma 5.39. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of complex numbers such that ak 6= 0, k ∈ {1, . . . , n} and
bk
bk
bk
bk
(5.97) m ≤ Re
+ Im
≤ M, m ≤ Re
− Im
≤ M,
ak
ak
ak
ak
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where m, M ∈ R and k ∈ {1, . . . , n} . Then one has the inequality
#
" n
n
n
X
X
X
(5.98)
|bk |2 + mM
|ak |2 ≤ (m + M ) Re
ak b̄k
k=1
k=1
k=1
n
X
≤ |M + m| ak b̄k .
k=1
Using the above result we may state and prove the following reverse inequality.
Theorem 5.40. If ā and b̄ are as in (5.97) and m, M > 0, then one has the
inequality
!2
n
n
n
X
X
X
(M + m)2
2
2
(5.99)
|ak |
|bk | ≤
ak b̄k
Re
4mM
k=1
k=1
k=1
n
2
(M + m)2 X
≤
a
b̄
k k .
4mM k=1
Proof. Using the elementary inequality
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αp + q 2 ≥ 2pq, α > 0, p, q ≥ 0
α
2
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we have
√
(5.100)
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n
X
1
mM
|ak | + √
mM
k=1
2
n
X
k=1
2
|bk | ≥ 2
n
X
n
X
2
|ak |
|bk |2
k=1
k=1
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! 12
.
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On the other hand, by (5.98), we have
(5.101)
#
" n
n
n
X
X
√
1 X
(M
+
m)
√
|bk |2 + mM
|ak |2 ≤ √
Re
ak b̄k
mM k=1
mM
k=1
n k=1
M + m X
≤ √
ak b̄k .
mM k=1
Combining (5.100) and (5.101), we deduce the desired result (5.99).
The following corollary is a natural consequence of the above lemma.
Corollary 5.41. If ā and b̄ and m, M satisfy the hypothesis of Theorem 5.40,
then
! 12 n
n
n
X
X
X
2
2
(5.102)
|ai |
|bi |
−
ai b̄i 0≤
i=1
i=1
i=1
1
!
!2
n
n
n
X
X
X
2
2
|ai |
|bi |
− Re
ak b̄k ≤
i=1
i=1
k=1
√
√ 2
!
n
M − m X
√
≤
Re
a
b̄
k k 2 mM
k=1
√
√ 2 n
M − m X
√
≤
ak b̄k 2 mM
k=1
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and
n
2
X
0≤
|ai |
|bi | − ai b̄i i=1
i=1
i=1
!2
n
n
n
X
X
X
≤
|ai |2
|bi |2 − Re
ai b̄i i=1
i=1
i=1
!2
n
X
(M − m)2 ai b̄i ≤
Re
4mM i=1
n
2
(M − m)2 X
≤
a
b̄
i i .
4mM n
X
(5.103)
(5.104)
(5.105)
(5.106)
2
n
X
2
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i=1
Another result obtained by Diaz and Metcalf in [19] is the following one.
Lemma 5.42. Let ā, b̄, m and M be complex numbers such that
bk
bk
(5.107) Re (m) + Im (m) ≤ Re
+ Im
≤ Re (M ) + Im (M ) ;
ak
ak
bk
bk
− Im
≤ Re (M ) − Im (M ) ;
Re (m) − Im (m) ≤ Re
ak
ak
(5.108)
k=1
2
|bk | + Re mM̄
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for each k ∈ {1, . . . , n} . Then
n
X
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n
X
k=1
"
2
|ak | ≤ Re (M + m)
n
X
k=1
#
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n
X
≤ |M + m| ak b̄k .
k=1
The following reverse result for the (CBS) −inequality may be stated as
well.
Theorem 5.43. With the assumptions in Lemma 5.42, and if Re mM̄ > 0,
then we have the inequality:
" n
# 12
P
n
X
X
Re (M + m) nk=1 ak b̄k
2
2
(5.109)
≤
|ak |
|bk |
1
2 Re mM̄ 2
k=1
k=1
P
|M + m| nk=1 ak b̄k ≤
.
12
2 Re mM̄
The proof is similar to the one in Theorem 5.40 and we omit the details.
Remark 5.8. Similar additive versions may be stated. They are left as an exercise for the interested reader.
5.15.
Some Reverse Results Via the Čebyšev Functional
For x̄= (x1 , . . . , xn ) , ȳ= (y1 , . . . , yn ) two sequences of real
Pnumbers and p̄ =
(p1 , . . . , pn ) a sequence of nonnegative real numbers with ni=1 pi = 1, define
the Čebyšev functional
(5.110)
Tn (p̄; x̄, ȳ) :=
n
X
i=1
pi xi yi −
n
X
i=1
pi xi
n
X
i=1
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pi yi .
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For x̄ and p̄ as above consider the norms:
kx̄k∞ := max |xi |
i=1,n
kx̄kp̄,α :=
n
X
! α1
pi |xi |α
, α ∈ [1, ∞).
i=1
The following result holds [20].
Theorem 5.44. Let x̄, ȳ, p̄ be as above and c̄= (c, . . . , c) a constant sequence
with c ∈ R. Then one has the inequalities
(5.111)
0 ≤ |Tn (p̄; x̄, ȳ)|

kȳ − ȳµ,p kp̄,1 · inf kx̄ − c̄k∞ ;


c∈R


kȳ − ȳµ,p kp̄,β · inf kx̄ − c̄kp̄,α , α > 1, α1 +
≤
c∈R



 kȳ − ȳ k · inf kx̄ − c̄k ;
µ,p ∞
c∈R
S.S. Dragomir
1
β
p̄,1

kȳ
−
ȳ
k
min
kx̄k
,
kx̄
−
x̄
k
;

µ,p
µ,p
∞
p̄,1
∞



n
o


kȳ − ȳµ,p kp̄,β min kx̄kp̄,α , kx̄ − x̄µ,p kp̄,α ,
≤

α > 1, α1 + β1 =o1;


n


 kȳ − ȳµ,p k · min kx̄k , kx̄ − x̄µ,p k
p̄,1
∞
p̄,1 ;
where
xµ,p :=
n
X
i=1
pi xi ,
yµ,p :=
n
X
i=1
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and x̄µ,p , ȳ µ,p are the sequences with all components equal to xµ,p , yµ,p .
Proof. Firstly, let us observe that for any c ∈ R, one has Sonin’s identity
(5.112)
Tn (p̄; x̄, ȳ) = Tn (p̄; x̄ − c̄, ȳ − ȳµ,p )
!
n
n
X
X
=
pi (xi − c) yi −
pj yj .
i=1
j=1
Taking the modulus and using Hölder’s inequality, we have
(5.113) |Tn (p̄; x̄, ȳ)| ≤
n
X
pi |xi − c| |yi − yµ,p |
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S.S. Dragomir
i=1

P

max |xi − c| ni=1 pi |yi − yµ,p |


i=1,n


P
1

1
 Pn
α α
n
β β
(
p
|x
−
c|
)
p
|y
−
y
|
,
i
i
i
i
µ,p
i=1
i=1
≤

1
1

α > 1, α + β = 1;

Pn



p
|x
−
c|
max
|y
−
y
i
µ,p |
 i=1 i i
i=1,n

kx̄ − c̄k∞ kȳ − ȳµ,p kp̄,1 ;



kx̄ − c̄kp̄,α kȳ − ȳµ,p kp̄,β , α > 1, α1 + β1 = 1;
=



kx̄ − c̄kp̄,1 kȳ − ȳµ,p k∞ .
Taking the inf over c ∈ R in (5.113), we deduce the second inequality in
(5.111).
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Since
inf kx̄ − c̄kp̄,α ≤
c∈R

 kx̄kp̄,α ,

for any α ∈ [1, ∞]
kx̄ − x̄µ,p kp̄,α
the last part of (5.110) is also proved.
For p̄ and x̄ as above, define
Tn (p̄; x̄) :=
n
X
pi x2i
i=1
−
n
X
!2
pi xi
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.
i=1
S.S. Dragomir
The following corollary holds [20].
Corollary 5.45. With the above assumptions we have
(5.114)
0 ≤ |Tn (p̄; x̄)|

kx̄ − x̄µ,p kp̄,1 · inf kx̄ − c̄k∞ ;


c∈R


kx̄ − x̄µ,p kp̄,β · inf kx̄ − c̄kp̄,α , α > 1, α1 +
≤
c∈R



 kx̄ − x̄ k · inf kx̄ − c̄k ;
µ,p ∞
c∈R
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1
β
p̄,1

kx̄
−
x̄
k
min
kx̄k
,
kx̄
−
x̄
k

µ,p
µ,p
∞
p̄,1
∞ ;



o
n


kx̄ − x̄µ,p kp̄,β min kx̄kp̄,α , kx̄ − x̄µ,p kp̄,α ,
≤

α > 1, α1 + β1 =o1;


n


 kx̄ − x̄µ,p k · min kx̄k , kx̄ − x̄µ,p k
p̄,1
p̄,1 .
∞
= 1;
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Remark 5.9. If pi := n1 , i = 1, . . . , n, then from Theorem 5.44 and Corollary
5.45 we recapture the results in [22].
The following reverse of the (CBS) −inequality holds [20].
Theorem 5.46. Let ā, b̄ be two sequences of real numbers with ai 6= 0, i ∈
{1, . . . , n} . Then one has the inequality
(5.115)
0≤
n
X
i=1
a2i
n
X
i=1
b2i −
n
X
!2
ai b i
i=1
n "
#
n
X
bi
X
a a |ai | ak k i ≤ inf max − c
bk bi c∈R i=1,n ai
i=1
k=1

bi 
"
#

max
n
n
X

X
ai ak ai i=1,n
×
Pn
≤
|ai | ak bi
ak bk bk bi 
k=1

i=1
k=1
.
 max − Pn
2 i=1,n ai
k=1 ak
Proof. By Corollary 5.45, we may state that
(5.116)
0 ≤ Tn (p̄; x̄)
≤ kx̄ − x̄µ,p kp̄,1 · inf kx̄ − c̄k∞
c∈R

 kx̄k∞ ,
≤ kx̄ − x̄µ,p kp̄,1 ×

kx̄ − x̄µ,p k∞ .
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For the choices
a2
pi = Pn i
2
k=1 ak
, xi =
bi
, i = 1, . . . , n;
ai
we get
Pn
Tn (p̄; x̄) =
kx̄ − x̄µ,p kp̄,1 =
=
=
=
=
i=1
a2i
Pn
Pn
2
i=1 bi − (
i=1
Pn
2 2
( k=1 ak )
,
n
X
pi xi −
pj xj i=1
j=1
n
n
b
X
X
1
1
i
2
Pn
P
a
−
a
b
j j
n
i
2
2
ai
k=1 ak i=1
k=1 ak j=1
n n
n
X
X
X
1
2
2
a
−
a
a
b
a
b
i i
Pn
j j
k
i
2 2
( k=1 ak ) i=1
j=1
k=1
n
n
X
X
1
|ai | ak (ak bi − ai bk )
Pn
2 2
( k=1 ak ) i=1
k=1 n
n
X
X
ak ai 1
,
|ai | ak P
2
bk bi ( nk=1 a2k )
n
X
i=1
kx̄ − c̄k∞
2
ai bi )
bi
= max − c ,
i=1,n ai
k=1
kx̄k∞
bi = max i=1,n ai
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and
kx̄ − x̄µ,p k∞
Pn
b
a
b
i
j=1 j j P
= max −
.
n
2 i=1,n ai
k=1 ak Utilising the inequality (5.116) we deduce the desired result (5.115).
The following result also holds [20].
Theorem 5.47. With the assumption in Theorem 5.46 and if α > 1,
then we have the inequality:
!2
n
n
n
X
X
X
(5.117) 0 ≤
a2i
b2i −
ai b i
i=1
i=1
1
α
+ β1 = 1,
i=1
S.S. Dragomir
 β1
n
β
n
X X
a a |ai |2−β ≤
ak k i 
bk bi 

i=1
c∈R
n
X
! α1
|ai |2−α |bi − cai |α
i=1
 β1
β
n
X
a a |ai |2−β ≤
ak k i 
bk bi i=1
k=1
 P
1
n
2−α
α α

|a
|
|b
|

i
i
i=1

1
×
ak ai α α
Pn
1

2−α Pn

 Pn
.
i=1 |ai |
2
k=1 ak bk bi a
k=1 k
n
X
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k=1
× inf

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Proof. By Corollary 5.45, we may state that
(5.118)
0 ≤ Tn (p̄; x̄)
≤ kx̄ − x̄µ,p kp̄,β · inf kx̄ − c̄kp̄,α
c∈R
(
kx̄kp̄,α ,
≤ kx̄ − x̄µ,p kp̄,β ×
kx̄ − x̄µ,p kp̄,α ,
for α > 1, α1 + β1 = 1.
For the choices
A Survey on
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a2
pi = Pn i
2
k=1 ak
, xi =
bi
, i = 1, . . . , n;
ai
S.S. Dragomir
Title Page
we get

kx̄ − x̄µ,p kp̄,β = 
n
X
i=1

=
n
X
Contents
β  β1
n
X
pi xi −
pj xj 
a2

Pn i
k=1
i=1
JJ
J
j=1
Pn
β  β1
Pn
2
bi
k=1 ak − ai
j=1 aj bj 
P
a2k ai nk=1 a2k

n
X
1

= P
1+ 1
( nk=1 a2k ) β i=1
1
n
β β
X a a |ai |2−β ak k i  ,
bk bi II
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kx̄ − c̄kp̄,α =
n
X
! α1
α
pi |xi − c|
i=1
n
X
1
= P
1
( nk=1 a2k ) α
! α1
2−α
|ai |
α
|bi − cai |
,
i=1
n
X
1
kx̄kp̄,α = P
1
( nk=1 a2k ) α
! α1
2−α
|ai |
α
|bi |
i=1
and
1
kx̄ − x̄µ,p kp̄,α = P
1+ 1
n
( k=1 a2k ) α
n
X
i=1
α ! 1
n
X
ak ai α
2−α |ai |
ak .
bk bi k=1
Utilising the inequality (5.118), we deduce the desired result (5.117).
Finally, the following result also holds [20].
Theorem 5.48. With the assumptions in Theorem 5.46 we have the following
reverse of the (CBS) −inequality:
!2
n
n
n
X
X
X
(5.119)
0≤
a2i
b2i −
ai b i
i=1
i=1
n
b X
i
a2k −
≤ max i=1,n ai
k=1
i=1
n
X
j=1
" n
#
X
aj bj inf
|ai | |bi − cai |
c∈R
i=1
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n
n
b X
X
i
2
≤ max ak −
aj b j i=1,n ai
j=1
k=1
 Pn

|a b |

 i=1 i i
×
Pn
ak ai Pn
1

.

 Pn
i=1 |ai | k=1 ak b
2
b
a
k
i
k=1 k
Proof. By Corollary 5.45, we may state that
A Survey on
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Type Discrete Inequalities
0 ≤ Tn (p̄; x̄)
≤ kx̄ − x̄µ,p k∞ · inf kx̄ − c̄kp̄,1
c∈R
 kx̄kp̄,1 ,
≤ kx̄ − x̄µ,p k∞
 kx̄ − x̄ k .
µ,p p̄,1
(5.120)
S.S. Dragomir
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For the choices
JJ
J
a2i
bi
pi = Pn
, xi = , i = 1, . . . , n;
2
ai
k=1 ak
Go Back
we get
kx̄ − x̄µ,p k∞
II
I
n
X
= max xi −
pj xj i=1,n j=1
P
n
n
n
b X
b
X
a
b
1
j
j
i
i
j=1
2
P
a
−
a
b
=
max
= max − Pn
j j ,
n
k
2
2
i=1,n ai
k=1 ak k=1 ak i=1,n ai
k=1
j=1
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kx̄ − c̄kp̄,1 =
n
X
pi |xi − c|
i=1
n
X
kx̄kp̄,1
n
X
bi
1
Pn
=
− c = Pn
|ai | |bi − cai | ,
2 2
a
a
a
i
k
k
k=1
k=1
i=1
i=1
n
n
n
X
X
X
bi 1
a2i
Pn
= Pn
|ai bi |
=
pi |xi | =
2
2 k=1 ak i=1
k=1 ak ai
i=1
i=1
a2i
and
1
kx̄ − x̄µ,p kp̄,1 = Pn
2
( k=1 a2k )
n
X
i=1
n
X a a |ai | ak k i .
bk bi k=1
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Utilising the inequality (5.120) we deduce (5.119).
Title Page
5.16.
Another Reverse Result via a Grüss Type Result
The following Grüss type inequality has been obtained in [21].
Lemma 5.49. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be two sequences of real
numbers and assume that there are γ, Γ ∈ R such that
(5.121)
−∞ < γ ≤ ai ≤ Γ < ∞ for each i ∈ {1, . . . , n} .
Then for any p̄ = (p1 , . . . , pn ) a nonnegative sequence with the property that
P
n
i=1 pi = 1, one has the inequality
n
n
n
n
n
X
1
X
X
X
X
(5.122) p i bi −
p i ai b i −
p i ai
pi bi ≤ (Γ − γ)
p k bk .
2
i=1
i=1
i=1
i=1
k=1
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The constant
constant.
1
2
is sharp in the sense that it cannot be replaced by a smaller
Proof. We will give here a simpler direct proof based on Sonin’s identity. A
simple calculation shows that:
n
X
(5.123)
p i ai b i −
i=1
n
X
p i ai
i=1
=
n
X
pi bi
i=1
n
X
i=1
pi
γ+Γ
ai −
2
bi −
n
X
!
p k bk
.
k=1
By (5.121) we have
ai − γ + Γ ≤ Γ − γ for all i ∈ {1, . . . , n}
2 2
S.S. Dragomir
Title Page
and thus, by (5.123), on taking the modulus, we get
n
n
n
n
n
X
X
X
X
X
γ
+
Γ
bi −
p i ai b i −
p i ai
p i bi ≤
pi ai −
p
b
k k
2
i=1
i=1
i=1
i=1
k=1
n
n
X
X
1
≤ (Γ − γ)
pi bi −
p k bk .
2
i=1
k=1
1
,
2
To prove the sharpness of the constant let us assume that (5.122) holds with
a constant c > 0, i.e.,
n
n
n
n
n
X
X
X
X
X
(5.124) p i ai b i −
p i ai
pi bi ≤ c (Γ − γ)
p i bi −
p k bk .
i=1
i=1
i=1
i=1
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provided ai satisfies (5.121).
If we choose n = 2 in (5.124) and take into account that
2
X
p i ai b i −
i=1
2
X
p i ai
i=1
2
X
pi bi = p1 p2 (a1 − a2 ) (b1 − b2 )
i=1
provided p1 + p2 = 1, p1 , p2 ∈ [0, 1] , and since
2
2
X
X
p i bi −
p k bk i=1
k=1
= p1 |(p1 + p2 ) b1 − p1 b1 − p2 b2 | + p2 |(p1 + p2 ) b2 − p1 b1 − p2 b2 |
= 2p1 p2 |b1 − b2 |
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S.S. Dragomir
we deduce by (5.124)
p1 p2 |a1 − a2 | |b1 − b2 | ≤ 2c (Γ − γ) |b1 − b2 | p1 p2 .
(5.125)
If we assume that p1 , p2 6= 0, b1 6= b2 and a1 = Γ, a2 = γ, then by (5.125) we
deduce c ≥ 21 , which proves the sharpness of the constant 21 .
The following corollary is a natural consequence of the above lemma.
Corollary 5.50. Assume that ā = (a1 , . . . , an ) satisfies the assumption (5.121)
and p̄ is a probability sequence. Then
!2
n
n
n
n
X
X
X
X
1
2
(5.126) 0 ≤
p i ai −
p i ai
≤ (Γ − γ)
p i ai −
pk ak .
2
i=1
The constant
1
2
i=1
i=1
is best possible in the sense mentioned above.
k=1
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The following reverse of the (CBS) −inequality may be stated.
Theorem 5.51. Assume that x̄ = (x1 , . . . , xn ) and ȳ = (y1 , . . . , yn ) are sequences of real numbers with yi 6= 0 (i = 1, . . . , n) . If there exists the real
numbers m, M such that
xi
≤ M for each i ∈ {1, . . . , n} ,
yi
m≤
(5.127)
then we have the inequality
0≤
(5.128)
n
X
x2i
i=1
≤
yi2 −
y2
Pn i
1
k=1
2
k=1 yk
n
X
yk2
!2
xi yi
n
X
i=1
n
X
x y |yi | yk · i i .
xk yk , ai =
k=1
xi
yi
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
i=1
1
(M − m)
2
Pn
n
X
i=1
Proof. If we choose pi =
in (5.126), we deduce
Pn 2
x
Pni=1 i2 −
k=1 yk
n
X
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for i = 1, . . . , n and γ = m, Γ = M
!2
xi yi
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i=1
n
X
1
1
1
2 xi
P
P
≤ (M − m) n
yi − n
xk yk 2
2
2
yi
k=1 yk i=1
k=1 yk k=1
n
n
n
X
X
X
1
1
2
= (M − m) Pn
|yi | xi
yk − yi
xk yk 2
2
2
( k=1 yk ) i=1
k=1
k=1
n
X
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n
n
X
X
xi yi 1
1
.
= (M − m) Pn
yk · |yi | 2
xk y k 2
( k=1 yk2 ) i=1
k=1
giving the desired inequality (5.128).
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Type Discrete Inequalities
S.S. Dragomir
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References
[1] R. FRUCHT, Sobre algunas disigualdades: Observación relative a la solución del Problema No. 21, indicada par el Ing. Ernesto M. Saleme (in
Spanish), Math. Notae, 3 (1943), 41–46.
[2] G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis,
Dept. of Experimental Statistics, North Carolina State College, Raleigh;
Univ. of North Carolina, Mimograph Ser., No. 49, 1951.
[3] G.S. WATSON, Serial correlation in regression analysis, I, Biometrika, 42
(1955), 327–341.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
[4] G.S. WATSON, A method for discovering Kantorovich-type inequalities
and a probabilistic interpretation, Linear Algebra Appl., 97 (1987), 211–
217.
S.S. Dragomir
[5] G.S. WATSON, G. ALPARGU and G.P.H. STYAN, Some comments on
six inequalities associated with the inefficiency of ordinary least squares
with one regressor, Linear Algebra and its Appl., 264 (1997), 13–54.
[6] G. PÓLYA and G. SZEGÖ, Aufgaben und Lehrsütze ans der Analysis,
Band I: Reihen, Integralrechnung, Funktiontheorie (in German), 4th Ed.,
Springer Verlag, Berlin, 1970 (Original version: Julius Springer, Berlin,
1925).
[7] G. PÓLYA and G. SZEGÖ, Problems and Theorems in Analysis, Volume
1: Series, Integral Calculus, Theory of Functions (in English), translated
from german by D. Aeppli, corrected printing of the revised translation of
the fourth German edition, Springer Verlag, New York, 1972.
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[8] W. GREUB AND W. RHEINBOLDT, On a generalisation of an inequality
of L.V. Kantorovich, Proc. Amer. Math. Soc., 10 (1959), 407–415.
[9] S.S. DRAGOMIR, A generalisation of Cassel’s and Greub-Reinboldt’s inequalities in inner product spaces, RGMIA Res. Rep. Coll., 6 (2003), Supplement. [ONLINE: http://rgmia.vu.edu.au/v6(E).html]
[10] S.S. DRAGOMIR, A counterpart of Schwartz’s inequality in inner product
spaces, RGMIA Res. Rep. Coll., 6 (2003), Supplement, Article 18. [ONLINE: http://rgmia.vu.edu.au/v6(E).html]
[11] S.S. DRAGOMIR AND G.L. BOOTH, Grüss-Lupaş type inequality and its
applications for the estimation of p-moments of guessing mappings, Math.
Comm., 5 (2000), 117–126.
[12] S.S. DRAGOMIR, Some counterpart inequalities in terms of forward difference for (CBS) −inequality, submitted.
[13] S.S. DRAGOMIR, A Grüss type inequality for sequences of vectors in
normed linear spaces and applications, submitted.
[14] S.S. DRAGOMIR, Another Grüss type inequality for sequences of vectors
in normed linear spaces and applications, J. Comp. Anal. & Applic., 4(2)
(2002), 155–172.
[15] D. ANDRICA AND C. BADEA, Grüss’ inequality for positive linear functionals, Periodica Mat. Hungarica, 19(2) (1988), 155–167.
[16] A.W. ROBERTS AND D.E. VARBERG, Convex Functions, Academic
Press, New York, 1973.
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Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
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[17] A. LUPAŞ, A remark on the Schweitzer and Kantorovich inequalities,
Publ. Elek. Fak. Univ. Beograde, Ser. Mat. i Fiz., 383 (1972), 13–14.
[18] P. SCHWEITZER, An inequality about the arithmetic mean (Hungarian),
Math. Phys. Lapok (Budapest), 23 (1914), 257–261.
[19] J.B. DIAZ AND F.T. METCALF, Stronger forms of a class of inequalities
of G. Pólya – G. Szegö and L.V. Kantorovich, Bull. Amer. Math. Soc., 69
(1963), 415–418.
[20] P. CERONE AND S.S. DRAGOMIR, New inequalities for the Čebyšev
functional involving two n−tuples of real numbers and applications,
RGMIA Res. Rep. Coll., 5(3) (2002), Article 4. [ONLINE: http://
rgmia.vu.edu.au/v5n3.html]
[21] P. CERONE AND S.S. DRAGOMIR, A refinement of the Grüss inequality
and applications, RGMIA Res. Rep. Coll., 5(2) (2002), Article 2. [ONLINE: http://rgmia.vu.edu.au/v5n2.html]
[22] S.S. DRAGOMIR, New inequalities for the weighted Čebyšev functionals
and applications for the (CBS) −inequality, in preparation.
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Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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I
[23] O. SHISHA AND B. MOND, Bounds on differences of means, Inequalities, Academic Press Inc., New York, 1967, 293–308.
Go Back
[24] D. ZAGIER, A converse to Cauchy’s inequality, Amer. Math. Month.,
102(10) (1995), 919–920.
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[25] D. ZAGIER, An inequality converse to that of Cauchy (Dutch), Indag.
Math., 39(4) (1997), 349–351.
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[26] J. PEČARIĆ, The inequalities of Zagier and Čebyšev, Arch. Math.,
(Basel), 64(5) (1995), 415–417.
[27] H. ALZER, On a converse Cauchy inequality of D. Zagier, Arch. Math.,
(Basel), 58(2) (1995), 157–159.
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Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
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6.
6.1.
Related Inequalities
Ostrowski’s Inequality for Real Sequences
In 1951, A.M. Ostrowski [2, p. 289] gave the following result related to the
(CBS) −inequality for real sequences (see also [1, p. 92]).
Theorem 6.1. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two non-proportional
sequences of real numbers. Let x̄ = (x1 , . . . , xn ) be a sequence of real numbers
such that
n
X
(6.1)
ai xi = 0 and
i=1
n
X
bi xi = 1.
S.S. Dragomir
i=1
Then
(6.2)
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n
X
x2i ≥
i=1
Pn
2
i=1 ai
Pn 2 Pn 2
Pn
i=1 ai
i=1 bi − (
i=1
Contents
bk
xk = Pn
i=1
Pn
i=1
a2i − ak
Pn
ai b i
Pi=1
2
n
2
i=1 bi − (
i=1 ai bi )
Pn
2
ai
for any k ∈ {1, . . . , n} .
2
ai b i )
with equality if and only if
(6.3)
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Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
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Proof. We shall follow the proof in [1, p. 93 – p. 94].
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Let
(6.4)
A=
n
X
a2i ,
B=
i=1
n
X
b2i ,
C=
i=1
n
X
ai b i
i=1
and
(6.5)
yi =
Abi − Cai
for any i ∈ {1, . . . , n} .
AB − C 2
It is easy to see that the sequence ȳ = (y1 , . . . , yn ) as defined by (6.5) satisfies
(6.1).
Any sequence x̄ = (x1 , . . . , xn ) that satisfies (6.1) fulfills the equality
n
X
xi yi =
n
X
i=1
xi ·
i=1
(Abi − Cai )
A
=
;
2
AB − C
AB − C 2
n
X
A
.
AB − C 2
yi2 =
i=1
Any sequence x̄ = (x1 , . . . , xn ) that satisfies (6.1) therefore satisfies
n
X
x2i −
i=1
and thus
S.S. Dragomir
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so, in particular
(6.6)
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
n
X
yi2 =
i=1
n
X
i=1
x2i
≥
n
X
(xi − yi )2 ≥ 0,
i=1
n
X
i=1
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yi2
A
=
AB − C 2
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and the inequality (6.2) is proved.
From (6.6) it follows that equality holds in (6.1) iff xi = yi for each i ∈
{1, . . . , n}, and the theorem is completely proved.
6.2.
Ostrowski’s Inequality for Complex Sequences
The following result that points out a natural generalisation of Ostrowski’s inequality for complex numbers holds [3].
Theorem 6.2. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) and x̄ = (x1 , . . . , xn )
be sequences of complex numbers. If ā and b̄, where b̄ = b̄1 , . . . , b̄n , are not
proportional and
n
X
(6.7)
xi āi = 0;
i=1
n
X
xi b̄i = 1,
(6.8)
i=1
then one has the inequality
(6.9)
n
X
i=1
2
Pn
|xi | ≥ Pn
i=1
|ai |2
Pn
2
2 Pn
2
|a
|
|b
|
−
a
b̄
i
i
i
i
i=1
i=1
i=1
S.S. Dragomir
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with equality iff
"
(6.10)
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Type Discrete Inequalities
#
Pn
b
ā
k
k
xi = µ bi − Pnk=1
2 · ai , i ∈ {1, . . . , n}
k=1 |ak |
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and µ ∈ C with
Pn
(6.11)
|µ| = Pn
k=1
k=1
|ak |2
Pn
k=1
|ak |2
P
2 .
|bk |2 − nk=1 ak b̄k Proof. Recall the (CBS) −inequality for complex sequences
2
n
n
n
X
X
X
2
2
(6.12)
|uk |
|vk | ≥ uk v̄k k=1
k=1
k=1
with equality iff there is a complex number α ∈ C such that
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
uk = αvk , k = 1, . . . , n.
S.S. Dragomir
Pn
zi c̄i
uk = zk − Pni=1 2 · ck ,
|ci |
Pi=1
n
di c̄i
vk = dk − Pni=1 2 · ck , where c̄ 6= 0 and c̄, d̄, z̄ ∈ Cn ,
i=1 |ci |
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(6.13)
If we apply (6.12) for
we have
(6.14)
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2
2
Pn
Pn
n n X
X
z
c̄
d
c̄
i
i
i
i
i=1
i=1
· ck · ck zk − Pn
dk − Pn
2
2
i=1 |ci |
i=1 |ci |
k=1
k=1
!
!2
Pn
Pn
n
X
z
c̄
d
c̄
i i
i i
≥
zk − Pni=1 2 · ck
dk − Pni=1 2 · ck i=1 |ci |
i=1 |ci |
k=1
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with equality iff there is a β ∈ C such that
!
Pn
Pn
z
c̄
d
c̄
i
i
i
i
(6.15)
zk = Pni=1 2 · ck + β dk − Pni=1 2 · ck .
i=1 |ci |
i=1 |ci |
Since a simple calculation shows that
2 P
Pn
Pn
n 2
n
2 Pn
2
X
z
c̄
k=1 |zk |
k=1 |ck | − |
k=1 zk c̄k |
i=1 i i
·
c
=
,
zk − Pn
k
Pn
2
2 2
|c
|
i
|c
|
i=1
k=1
k=1 k
2 P
P
P
Pn
n
2
n
n
2
n
2
X i=1 di c̄i
k=1 |dk |
k=1 |ck | − |
k=1 dk c̄k |
dk − Pn
Pn
2 · ck =
2 2
|c
|
i=1 |ci |
k
k=1
k=1
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
and
n
X
k=1
!
!
Pn
Pn
z
c̄
d
c̄
i i
i i
zk − Pni=1 2 · ck
dk − Pni=1 2 · ck
i=1 |ci |
i=1 |ci |
Pn
P
Pn
Pn
n
2
¯
¯
k=1 zk dk ·
k=1 |ck | −
k=1 zk c̄k ·
k=1 ck dk
=
Pn
2 2
i=1 |ci |
then by (6.12) we deduce

2 
n
n
n
X
X
X
(6.16) 
|zk |2
|ck |2 − zk c̄k 
k=1
k=1
k=1

2 
n
n
n
X
X
X
2
2

×
|dk |
|ck | − dk c̄k 
k=1
k=1
k=1
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2
n
n
n
n
X
X
X
X
|ck |2 −
zk c̄k
ck d¯k ≥
zk d¯k ·
k=1
k=1
k=1
k=1
with equality iff there is a β ∈ C such that (6.15) holds.
If ā, x̄, b̄ satisfy (6.7) and (6.8), then by (6.16) and (6.15) for the choices
z̄ = x̄, c̄ = ā and d̄ = b̄, we deduce (6.9) with equality iff there is a µ ∈ C
such that
!
Pn
b̄
a
i i
xk = µ bk − Pni=1 2 · ak ,
i=1 |ai |
and, by (6.8),
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
S.S. Dragomir
!
Pn
n
X
i=1 ai b̄i
· ak · b̄k = 1.
bk − Pn
µ
2
i=1 |ai |
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Since (6.17) is clearly equivalent to (6.15), the theorem is completely proved.
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(6.17)
k=1
6.3.
Another Ostrowski’s Inequality
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In his book from 1951, [2, p. 130], A.M. Ostrowski proved the following inequality as well (see also [1, p. 94]).
Theorem 6.3. Let ā, b̄, x̄ be sequences of real numbers so that ā 6= 0 and
(6.18)
n
X
k=1
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x2k
=1
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n
X
(6.19)
ak xk = 0.
k=1
Then
Pn
(6.20)
k=1
a2k
Pn
2
Pn
b2 − (
k=1
Pnn 2
k=1 ak
k=1
ak b k )
≥
n
X
!2
bk x k
.
k=1
If ā and b̄ are non-proportional, then equality holds in (6.20) iff
(6.21)
P
P
bk ni=1 a2i − ak ni=1 ai bi
xk = q ·
hP
i1 ,
1
Pn
Pn 2
Pn
2 2
n
2
2 2
( k=1 ak )
i=1 ai
i=1 bi − (
i=1 ai bi )
k ∈ {1, . . . , n} , q ∈ {−1, 1} .
We may extend this result for sequences of complex numbers as follows [4].
Theorem 6.4. Let ā, b̄, x̄ be sequences of complex numbers so that ā 6= 0, ā, b̄
are not proportional, and
(6.22)
(6.23)
n
X
k=1
n
X
k=1
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S.S. Dragomir
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|xk | = 1
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xk āk = 0.
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Then
Pn
(6.24)
2
k=1 |ak |
2
Pn
2 n
2
X
|b
|
−
a
b̄
k=1 k
k=1 k k
≥
xk b̄k .
Pn
2
k=1 |ak |
k=1
Pn
The equality holds in (6.24) iff
!
Pn
bi āi
(6.25)
xk = β bk − Pni=1 2 · ak ,
i=1 |ai |
k ∈ {1, . . . , n} ;
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
where β ∈ C is such that
Pn
(6.26)
|β| = Pn
k=1
2
k=1 |ak |
2
|ak |
21
S.S. Dragomir
Pn
2 21
|b
|
−
a
b̄
k=1 k
k=1 k k
Pn
.
2
Title Page
Proof. In Subsection 6.2, we proved the following inequality:

2 
n
n
n
X
X
X
(6.27) 
|zk |2
|ck |2 − zk c̄k 
k=1
k=1
k=1

2 
n
n
n
X
X
X
×
|dk |2
|ck |2 − dk c̄k 
k=1
k=1
k=1
2
n
n
n
n
X
X
X
X
≥
zk d¯k ·
|ck |2 −
zk c̄k
ck d¯k k=1
k=1
k=1
k=1
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for any z̄, c̄, d̄ sequences of complex numbers, with equality iff there is a β ∈ C
such that
!
Pn
Pn
z
c̄
d
c̄
i
i
i
i
(6.28)
zk = Pni=1 2 · ck + β dk − Pni=1 2 · ck
i=1 |ci |
i=1 |ci |
for each k ∈ {1, . . . , n} .
If in (6.27) we choose z̄ = x̄, c̄ = ā and d̄ = b̄ and take into consideration
that (6.22) and (6.23) hold, then we get

2 
n
n
n
n
n
X
X
X
X
X
2
2
2
2
|xk |
|ak |
|bk |
|ak | − bk āk 
k=1
k=1
k=1
k=1
k=1
2
!2
n
n
X
X
≥
xk b̄k |ak |2
k=1
which is clearly equivalent to (6.24).
By (6.28) the equality holds in (6.24) iff
!
Pn
b
ā
i i
xk = β bk − Pni=1 2 · ak ,
i=1 |ai |
k=1
k ∈ {1, . . . , n} .
Since x̄ should satisfy (6.22), we get
2
" n
#
Pn
n
n X
X
X 2 |Pn bk āk |2
b
ā
i
i
2
2
2
1=
|xk | = |β|
|bk | − Pk=1
bk − Pni=1 2 · ak = |β|
n
2
|a
|
i
i=1
k=1 |ak |
k=1
k=1
k=1
from where we deduce that β satisfies (6.26).
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6.4.
Fan and Todd Inequalities
In 1955, K. Fan and J. Todd [5] proved the following inequality (see also [1, p.
94]).
Theorem 6.5. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be two sequences of
real numbers such that ai bj 6= aj bi for i 6= j. Then
Pn 2
a
(6.29) Pn 2 Pn i=12 i Pn
2
( i=1 ai ) ( i=1 bi ) − ( i=1 ai bi )
2

X
n
n
aj
n

X
≤

 .
2
a
b
−
a
b
j
i
i
j
j=1
i=1
A Survey on
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Type Discrete Inequalities
S.S. Dragomir
j6=i
Proof. We shall follow the proof in [1, p. 94 – p. 95].
Define
−1 X
aj
n
xi :=
(1 ≤ i ≤ n) .
2
aj b i − ai b j
j6=i
The terms in the sum on the right-hand side


n
n
n
−1
X
X X
ai aj
n

x i ai =


2
aj bi − ai bj
j=1
i=1
i=1
j6=i
can be grouped in pairs of the form
−1 ai aj
aj ai
n
+
2
aj b i − ai b j ai b j − aj b i
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and the sum of each such pair vanishes.
Hence, we deduce
n
X
i=1
ai xi = 0 and
n
X
bi xi = 1.
i=1
Applying Ostrowski’s inequality (see Section 6.1) we deduce the desired result
(6.29).
A weighted version of the result is also due to K. Fan and J. Todd [5] (see
also [1, p. 95]). We may state the result as follows.
Theorem 6.6. Let pij (i, j ∈ {1, . . . , n} , i 6= j) be real numbers such that
(6.30)
pij = pji ,
P
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S.S. Dragomir
for any i, j ∈ {1, . . . , n} with i 6= j.
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Denote P := 1≤i<j≤n pij and assume that P 6= 0. Then for any two sequences
of real numbers ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) satisfying ai bj 6= aj bi
(i 6= j) , we have

2
Pn 2
n
n
ai
1 X X pij aj 
(6.31) Pn 2 Pn i=1
≤
Pn

 .
2
2
2
P
a
b
−
a
b
a
b
−
(
a
b
)
j
i
i
j
i
i
j=1
i=1 i
i=1 i
i=1
i=1
j6=i
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6.5.
Some Results for Asynchronous Sequences
If S (R) is the linear space of real sequences, S+ (R) is the subset of nonnegative
sequences and Pf (N) denotes the set of finite parts of N, then for the functional
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T : Pf (N) × S+ (R) × S 2 (R) → R,
! 12
(6.32)
X
T I, p̄, ā, b̄ :=
pi a2i
X
i∈I
pi b2i
X
−
p i ai b i i∈I
i∈I
we may state the following result [6, Theorem 3].
Theorem 6.7. If |ā| = (|ai |)i∈N and b̄ = (|bi |)i∈N are asynchronous, i.e.,
A Survey on
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Type Discrete Inequalities
(|ai | − |aj |) (|bi | − |bj |) ≤ 0
for all i, j ∈ N, then
S.S. Dragomir
P
(6.33)
i∈I
T I, p̄, ā, b̄ ≥
pi |ai |
P
P
i∈I
i∈I
pi |bi |
pi
−
X
pi |ai bi | ≥ 0.
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i∈I
Contents
Proof. We shall follow the proof in [6].
Consider the inequalities
JJ
J
! 12
X
pi
i∈I
X
pi a2i
≥
i∈I
X
pi |ai |
II
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and
! 12
X
i∈I
pi
X
i∈I
pi b2i
≥
X
pi |bi |
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which by multiplication give
! 21
X
i∈I
pi a2i
X
pi b2i
i∈I
P
≥
i∈I
pi |ai |
P
P
i∈I
i∈I
pi
pi |bi |
.
Now, by the definition of T and by Čebyšev’s inequality for asynchronous sequences, we have
P
P
X
i∈I pi |ai |
i∈I pi |bi |
P
T I, p̄, ā, b̄ ≥
−
p i ai b i i∈I pi
i∈I
P
P
pi |ai | i∈I pi |bi | X
≥ i∈I P
−
pi |ai | |bi | ≥ 0
i∈I pi
i∈I
and the theorem is proved.
The following result also holds [6, Theorem 4].
Theorem 6.8. If |ā| and b̄ are synchronous, i.e., (|ai | − |aj |) (|bi | − |bj |) ≥ 0
for all i, j ∈ N, then one has the inequality
(6.34)
0 ≤ T I, p̄, ā, b̄ ≤ T I, p̄, āb̄, 1 ,
where 1 = (ei )i∈N , ei = 1, i ∈ N.
Proof. We have, by Čebyšev’s inequality for the synchronous sequences ā2 =
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(a2i )i∈N and b̄2 = (b2i )i∈N , that
! 12
T I, p̄, ā, b̄ =
≤
X
2
2
p i bi
p i ai
−
p i ai b i i∈I
i∈I
i∈I
! 21 X
X
X
pi a2i b2i
pi
−
pi ai bi = T I, p̄, āb̄, 1
X
i∈I
X
i∈I
i∈I
and the theorem is proved.
6.6.
A Survey on
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Type Discrete Inequalities
An Inequality via A − G − H Mean Inequality
S.S. Dragomir
The following result holds [6, Theorem 5].
Theorem 6.9. Let ā and b̄ be sequences of positive real numbers. Define
2
ai
b2i
(6.35)
∆i = P
P
2
2
ai
bi i∈I
i∈I
where i ∈ I and I is a finite part of N. Then one has the inequality
" # P 21P 2
2
P
Y ai ∆i i∈I ai i∈I bi
a
b
i i
P i∈I2 P
(6.36)
≥
2
bi
i∈I ai
i∈I bi
i∈I
P
P
2
b2
i∈I ai
≥ P a3 Pi∈I b3i .
i
i
i∈I bi
i∈I ai
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The equality holds in all the inequalities from (6.36) iff there exists a positive
number k > 0 such that ai = kbi for all i ∈ I.
Proof. We shall follow the proof in [6].
We will use the AGH−inequality
(6.37)
1 X
pi xi ≥
PI i∈I
! P1
Y
I
xpi i
≥P
PI
pi
i∈I xi
i∈I
,
P
where pi > 0, xi ≥ 0 for all i ∈ I, where PI := i∈I pi > 0.
We remark that the equality holds in (6.37) iff xi = xj for each i, j ∈ I.
Choosing pi = a2i and xi = abii (i ∈ I) in (6.37), then we get
(6.38)
P
P
P a2i 2
2
Y
a
b
bi
i∈I ai
i∈I i i
i∈I ai
P
≥ P a3
≥
2
i
ai
i∈I ai
i∈I
i∈I bi
and by pi = b2i and xi =
(6.39)
ai
,
bi
we also have
P
P
P b2i 2
2
Y
a
b
ai
i∈I bi
i∈I i i
i∈I bi
P
≥
≥
P b3i .
2
b
i
i∈I bi
i∈I
i∈I ai
If we multiply (6.38) with (6.39) we easily deduce the desired inequality (6.36).
The case of equality follows by the same case in the arithmetic mean – geometric mean – harmonic mean inequality. We omit the details.
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S.S. Dragomir
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The following corollary holds [6, Corollary 5.1].
Corollary 6.10. With ā and b̄ as above, one has the inequality
P
(6.40)
b3i
i∈I ai
2
P
i∈I ai bi
a3i
i∈I bi

P
 12
 ≥
P
2
i∈I bi
2 .
P
i∈I ai bi
i∈I
a2i
P
The equality holds in (6.40) iff there is a k > 0 such that ai = kbi , i ∈
{1, . . . , n} .
6.7.
A Related Result via Jensen’s Inequality for Power
Functions
The following result also holds [6, Theorem 6].
Theorem 6.11. Let ā and b̄ be sequences of positive real numbers and p ≥ 1.
If I is a finite part of N, then one has the inequality
"P
#1
2
2−p p P
p 2−p p
a
b
a
b
a
b
i i
i
i∈I i
P i∈I2 P
P
Pi∈I 2i i
.
≤
2
2
i∈I ai
i∈I bi
i∈I ai
i∈I bi
P
(6.41)
The equality holds in (6.41) if and only if there exists a k > 0 such that ai = kbi
for all i ∈ I.
If p ∈ (0, 1) , the inequality in (6.41) reverses.
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Proof. We shall follow the proof in [6].
By Jensen’s inequality for the convex mapping f : R+ → R,
f (x) = xp , p ≥ 1
one has
P
i∈I
(6.42)
pi xi
PI
p
P
≤
i∈I
pi xpi
PI
,
P
where PI := i∈I pi , pi > 0, xi ≥ 0, i ∈ I. The equality holds in (6.42) iff
xi = xj for all i, j ∈ I.
Now, choosing in (6.42) pi = a2i , xi = abii , we get
! p1
P
P
2−p p
bi
i∈I ai bi
i∈I ai
P
P
(6.43)
≤
2
2
i∈I ai
i∈I ai
and for pi = b2i , xi =
(6.44)
ai
,
bi
the inequality (6.42) also gives
! p1
P
P
p 2−p
a
b
a
b
i i
i∈I i i
Pi∈I 2 ≤
P
.
2
i∈I bi
i∈I bi
By multiplying the inequalities (6.43) and (6.44), we deduce the desired result
from (6.42).
The case of equality follows by the fact that in (6.42) the equality holds iff
(xi )i∈I is constant.
If p ∈ (0, 1) , then a reverse inequality holds in (6.42) giving the corresponding result in (6.41).
Remark 6.1. If p = 2, then (6.41) becomes the (CBS) −inequality.
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S.S. Dragomir
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6.8.
Inequalities Derived from the Double Sums Case
Let A = (aij )i,j=1,n and B = (bij )i,j=1,n be two matrices of real numbers. The
following inequality is known as the (CBS) −inequality for double sums
n
X
(6.45)
!2
aij bij
≤
i,j=1
n
X
a2ij
i,j=1
n
X
b2ij
i,j=1
with equality iff there is a real number r such that aij = rbij for any i, j ∈
{1, . . . , n} .
The following inequality holds [7, Theorem 5.2].
A Survey on
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S.S. Dragomir
Theorem 6.12. Let ā = (a1 , . . . , an ) and b̄ = (b1 , . . . , bn ) be sequences of real
numbers. Then
!2
!2
n
n
n
X
X
X
(6.46) ak +
bk − 2n
ak bk k=1
k=1
k=1
≤n
n
X
a2k
+
b2k
−2
k=1
n
X
ak
k=1
n
X
bk .
k=1
Proof. We shall follow the proof from [7].
Applying (6.45) for aij = ai − bj , bij = bi − aj and taking into account that
n
X
i,j=1
(ai − bj ) (bi − aj ) = 2n
n
X
k=1
ak b k −
n
X
k=1
!2
ak
−
n
X
k=1
!2
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n
X
(ai − bj )2 = n
i,j=1
and
n
X
n
X
n
n
X
X
bk
a2k + b2k − 2
ak
k=1
2
(bi − aj ) = n
i,j=1
n
X
a2k
+
b2k
−2
k=1
k=1
n
X
n
X
ak
k=1
k=1
bk ,
k=1
we may deduce the desired inequality (6.46).
The following result also holds [7, Theorem 5.3].
Theorem 6.13. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) , c̄ = (c1 , . . . , cn ) and
d̄ = (d1 , . . . , dn ) be sequences of real numbers. Then one has the inequality:

 Pn
i=1
(6.47)
det 
Pn
Pn
ai c i
i=1 bi ci
i=1
Pn
 Pi=1
n
ai di
Pn
Contents
bi di
i=1
Pn
a2i
i=1
ai b i
ai b i
Pn 2
bi
i=1Pn
× det  P
n
JJ
J


2
i=1 ci
i=1 ci di
Proof. We shall follow the proof in [7].
S.S. Dragomir
Title Page

i=1
≤ det 
2
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Pn
i=1 ci di
Pn
2
i=1 di
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Applying (6.45) for aij = ai bj − aj bi , bij = ci dj − cj di and using CauchyBinet’s identity [1, p. 85]
n
1X
(6.48)
(ai bj − aj bi ) (ci dj − cj di )
2 i,j=1
=
n
X
i=1
ai c i
n
X
bi di −
i=1
n
X
!
n
X
ai di
i=1
!
bi c i
i=1
A Survey on
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Type Discrete Inequalities
and Lagrange’s identity [1, p. 84]
(6.49)
n
n
n
X
X
1X
(ai bj − aj bi )2 =
a2i
b2i −
2 i,j=1
i=1
i=1
n
X
!2
ai b i
S.S. Dragomir
,
i=1
Title Page
we deduce the desired result (6.47).
6.9.
A Functional Generalisation for Double Sums
The following result holds [7, Theorem 5.5].
Theorem 6.14. Let A be a subset of real numbers R, f : A → R and ā =
(a1 , . . . , an ), b̄ = (b1 , . . . , bn ) sequences of real numbers with the property that
(i) ak bi , a2i , b2k ∈ A for any i, k ∈ {1, . . . , n} ;
(ii) f (a2k ) , f (b2k ) ≥ 0 for any k ∈ {1, . . . , n} ;
(iii) f 2 (ak bi ) ≤ f (a2k ) f (b2i ) for any i, k ∈ {1, . . . , n} .
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Then one has the inequality
" n
#2
n
n
X
X
X
2
2
f b2k .
f ak
(6.50)
f (ak bi ) ≤ n
k=1
k,i=1
k=1
Proof. We will follow the proof in [7].
Using the assumption (iii) and the (CBS) −inequality for double sums, we
have
n
n
X
X
(6.51)
f (ak bi ) ≤
|f (ak bi )|
k,i=1
≤
k,i=1
n
X
1
f a2k f b2i 2

!2
n
 X
1
≤
f a2k 2

n
X
f b2i
k,i=1
n
X
f a2k
f b2i
k,i=1
k,i=1
"
=n
n
X
# 12 "
f a2k
k=1
which is clearly equivalent to (6.50).
!2  12
1 
2

k,i=1
n
X
=
S.S. Dragomir
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k,i=1
"
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# 12
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X
k=1
# 12
f b2k
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The following corollary is a natural consequence of the above theorem [7,
Corollary 5.6].
Corollary 6.15. Let A, f and ā be as above. If
(i) ak ai ∈ A for any i, k ∈ {1, . . . , n} ;
(ii) f (a2k ) ≥ 0 for any k ∈ {1, . . . , n} ;
(iii) f 2 (ak ai ) ≤ f (a2k ) f (a2i ) for any i, k ∈ {1, . . . , n} ,
A Survey on
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then one has the inequality
n
n
X
X
(6.52)
f (ak ai ) ≤ n
f a2k .
k,i=1
S.S. Dragomir
k=1
The following particular inequalities also hold [7, p. 23].
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1. If ϕ : N → N is Euler’s indicator and s (n) denotes the sum of all relatively
prime numbers including and less than n, then for any ā = (a1 , . . . , an ),
b̄ = (b1 , . . . , bn ) sequences of natural numbers, one has the inequalities
" n
#2
n
n
X
X
X
2
2
(6.53)
ϕ (ak bi ) ≤ n
ϕ ak
ϕ b2k ;
(6.54)
k,i=1
n
X
ϕ (ak ai ) ≤ n
k,i=1
"
(6.55)
n
X
k,i=1
k=1
n
X
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k=1
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ϕ a2k ;
Quit
k=1
#2
s (ak bi )
≤n
2
n
X
k=1
s
a2k
n
X
k=1
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;
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n
X
(6.56)
k,i=1
s (ak ai ) ≤ n
n
X
s a2k .
k=1
2. If a > 1 and ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) are sequences of real
numbers, then
" n
#2
n
n
X
X
X
2
2
(6.57)
expa (ak bi ) ≤ n
expa ak
expa b2k ;
(6.58)
k,i=1
n
X
k,i=1
k=1
expa (ak ai ) ≤ n
n
X
k=1
expa a2k ;
k=1
3. If ā and b̄ are sequences of real numbers such that ak , bk ∈ (−1, 1)
(k ∈ {1, . . . , n}) , then one has the inequalities:
" n
#2
n
n
X
X
X
1
1
1
2
(6.59)
≤n
m,
m
2 m
(1 − ak bi )
(1 − ak ) k=1 (1 − b2k )
k,i=1
k=1
(6.60)
n
X
n
X
1
1
,
m ≤ n
2 m
(1
−
a
(1
−
a
)
k ai )
k
k,i=1
k=1
where m > 0.
6.10.
A (CBS) −Type Result for Lipschitzian Functions
The following result was obtained in [8, Theorem].
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S.S. Dragomir
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Theorem 6.16. Let f : I ⊆ R → R be a Lipschitzian function with the constant
M, i.e., it satisfies the condition
(6.61)
|f (x) − f (y)| ≤ M |x − y| for any x, y ∈ I.
If ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) are sequences of real numbers with ai bj ∈
I for any i, j ∈ {1, . . . , n} , then
n
n
X
X
(6.62)
0≤
f (ai bj ) |f (ai bj )| −
|f (aj bi )| f (ai bj )
≤
i,j=1
n
X
n
X
i,j=1
i,j=1
i,j=1
f 2 (aj bi ) −
f (ai bj ) f (aj bi )

n
n
X
X
2
2
≤M
ai
b2i −
i=1
i=1
n
X
!2 
ai b i
.
i=1
Proof. We shall follow the proof in [8].
Since f is Lipschitzian with the constant M, we have
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(6.63) 0 ≤ ||f (ai bj )| − |f (aj bi )|| ≤ |f (ai bj ) − f (aj bi )| ≤ M |ai bj − aj bi |
for any i, j ∈ {1, . . . , n} , giving
(6.64)
0 ≤ |(|f (ai bj )| − |f (aj bi )|) (f (ai bj ) − f (aj bi ))|
≤ (f (ai bj ) − f (aj bi ))2 ≤ M 2 (ai bj − aj bi )2
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for any i, j ∈ {1, . . . , n} .
The inequality (6.64) is obviously equivalent to
|f (ai bj )| f (ai bj ) + |f (aj bi )| f (aj bi )
(6.65)
− |f (ai bj )| f (ai bj ) − |f (aj bi )| f (aj bi ) ≤ f 2 (ai bj ) − 2f (ai bj ) f (aj bi ) + f 2 (aj bi )
≤ M 2 a2i b2j − 2ai bi aj bj + a2j b2i
for any i, j ∈ {1, . . . , n} . Summing over i and j from 1 to n in (6.65) and taking
into account that:
n
X
i,j=1
n
X
|f (ai bj )| f (ai bj ) =
|f (ai bj )| f (aj bi ) =
i,j=1
n
X
2
f (ai bj ) =
i,j=1
n
X
i,j=1
n
X
i,j=1
n
X
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|f (aj bi )| f (ai bj ) ,
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2
f (aj bi ) ,
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we deduce the desired inequality.
The following particular inequalities hold [8, p. 27 – p. 28].
1. Let x̄ = (x1 , . . . , xn ) , ȳ = (y1 , . . . , yn ) be sequences of real numbers
such that 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} . Then for any
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r ≥ 1 one has
(6.66)
0≤
n
X
x2r
i
i=1
n
X
n
X
yi2r −
!2
|xi yi |r
i=1
 i=1
n
n
X
X
≤ r2 (M1 M2 )2(r−1) 
x2i
yi2 −
i=1
i=1
n
X
!2 
|xi yi |
.
i=1
2. If 0 < m1 ≤ |xi | , 0 < m2 ≤ |yi | , i ∈ {1, . . . , n} and r ∈ (0, 1) , then
(6.67)
0≤
n
X
x2r
i
i=1
≤
n
X
yi2r −
i=1
r
2
(m1 m2 )2(r−1)
n
X
!2
i=1
S.S. Dragomir
|xi yi |r
i=1

n
n
X X
2

xi
yi2 −
i=1
n
X
!2 
|xi yi |
.
i=1
3. If 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} , then for any natural
number k one has
n
n
X
X
2k+1
2k+1
(6.68) 0 ≤ xi
|xi |
yi2k+1 |yi |2k+1
i=1
i=1
n
n
X
X
2k+1
2k+1 2k+1
−
x2k+1
|y
|
y
|x
|
i
i
i
i
i=1
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≤
n
X
n
X
2(2k+1)
xi
i=1
2(2k+1)
yi
−
n
X
!2
x2k+1
yi2k+1
i
i=1
i=1

n
n
X X
2
4k 
2
≤ (2k + 1) (M1 M2 )
xi
yi2 −
i=1
i=1
n
X
!2 
|xi yi |
.
i=1
4. If 0 < m1 ≤ xi , 0 < m2 ≤ yi , for any i ∈ {1, . . . , n} , then one has the
inequality
2 "X
# 2
n
n X
xi
xi
−
ln
ln
(6.69)
0≤n
yi
yi
i=1
i=1

!2 
n
n
n
X
X
X
1

x2i
yi2 −
≤
xi yi  .
2
(m1 m2 ) i=1
i=1
i=1
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6.11.
An Inequality via Jensen’s Discrete Inequality
The following result holds [9].
Theorem 6.17. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be two sequences of real
numbers with ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a convex (concave)
function on I and abii ∈ I for each i ∈ {1, . . . , n} , and w̄ = (w1 , . . . , wn ) is a
sequence of nonnegative real numbers, then
Pn
bi
Pn
2
w
a
f
i=1 i i
ai
wi ai bi
i=1
Pn
(6.70)
f Pn
≤ (≥)
.
2
2
i=1 wi ai
i=1 wi ai
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Proof. We shall use Jensen’s discrete inequality for convex (concave) functions
f
(6.71)
n
1 X
pi xi
Pn i=1
!
n
1 X
≤ (≥)
pi f (xi ) ,
Pn i=1
P
where pi ≥ 0 with Pn := ni=1 pi > 0 and xi ∈ I for each i ∈ {1, . . . , n} .
If in (6.71) we choose xi = abii and pi = wi a2i , then by (6.71) we deduce
the desired result (6.70).
The following corollary holds [9].
Corollary 6.18. Let ā and b̄ be sequences of positive real numbers and assume
that w̄ is as above. If p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the
inequality
!p
!p−1 n
n
n
X
X
X
(6.72)
wi ai bi
≤ (≥)
wi a2i
wi a2−p
bpi .
i
i=1
i=1
i=1
Proof. Follows by Theorem 6.17 applied for convex (concave) function f :
[0, ∞) → R, f (x) = xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) .
Remark 6.2. If p = 2, then by (6.72) we deduce the (CBS) −inequality.
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6.12.
An Inequality via Lah-Ribarić Inequality
The following reverse of Jensen’s discrete inequality was obtained in 1973 by
Lah and Ribarić [10].
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Lemma 6.19. Let f : I → R be a convex function, xi ∈ [m, M ] ⊆ I for each
i ∈ {1, . . . , n} and p̄ = (p1 , . . . , pn ) be a positive n−tuple. Then
(6.73)
n
1 X
pi f (xi )
Pn i=1
≤
M−
1
Pn
Pn
i=1 pi xi
M −m
f (m) +
1
Pn
Pn
i=1
pi xi − m
M −m
f (M ) .
Proof. We observe for each i ∈ {1, . . . , n} , that
(6.74)
xi =
(M − xi ) m + (xi − m) M
.
M −m
If in the definition of convexity, i.e., α, β ≥ 0, α + β > 0
αf (a) + βf (b)
αa + βb
(6.75)
f
≤
α+β
α+β
we choose α = M − xi , β = xi − m, a = m and b = M, we deduce, by (6.75),
that
(M − xi ) m + (xi − m) M
(6.76)
f (xi ) = f
M −m
(M − xi ) f (m) + (xi − m) f (M )
≤
M −m
for each i ∈ {1, . . . , n} .
If we multiply (6.76) by pi > 0 and sum over i from 1 to n, we deduce
(6.73).
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The following result holds.
Theorem 6.20. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be two sequences
of real numbers with ai 6= 0, i ∈ {1, . . . , n} . If I ⊆ R → R is a convex
(concave) function on I and abii ∈ [m, M ] ⊆ I for each i ∈ {1, . . . , n} and
w̄ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then
Pn
(6.77)
bi
2
i=1 wi ai f ai
Pn
2
i=1 wi ai
M−
≤ (≥)
Pn
wi ai bi
Pi=1
n
2
i=1 wi ai
M −m
f (m) +
Pn
wi ai bi
Pi=1
n
2
i=1 wi ai
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−m
M −m
f (M ) .
Proof. Follows by Lemma 6.19 for the choices pi = wi a2i , xi =
{1, . . . , n} .
bi
,
ai
S.S. Dragomir
i ∈
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The following corollary holds.
Corollary 6.21. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be two sequences of
positive real numbers and such that
(6.78)
0<m≤
bi
≤ M < ∞ for each i ∈ {1, . . . , n} .
ai
If w̄ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers and p ∈
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(−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequality
(6.79)
n
X
n
wi a2−p
bpi
i
i=1
M m (M p−1 − mp−1 ) X
+
wi a2i
M −m
i=1
n
M p − mp X
≤ (≥)
wi ai bi .
M − m i=1
Proof. If we write the inequality (6.77) for the convex (concave) function f (x) =
xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , we get
n
wi ai bi
Pn
2−p p
Pi=1
M
−
n
2
w
a
b
i
i
i
i=1 wi ai
i=1
P
≤
(≥)
· mp +
n
2
M
−
m
w
a
i
i
i=1
P
Pn
wi ai bi
Pi=1
n
2
i=1 wi ai
−m
M −m
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· M p,
which, after elementary calculations, is equivalent to (6.79).
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Remark 6.3. For p = 2, we get
(6.80)
n
X
i=1
wi b2i + M m
n
X
i=1
wi a2i ≤ (M + m)
n
X
wi ai bi ,
i=1
which is the well known Diaz-Metcalf inequality [11].
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6.13.
An Inequality via Dragomir-Ionescu Inequality
The following reverse of Jensen’s inequality was proved in 1994 by S.S. Dragomir
and N.M. Ionescu [12].
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Lemma 6.22. Let f : I ⊆ R → R be a differentiable convex function
Pnon I̊, xi ∈I̊
(i ∈ {1, . . . , n}) and pi ≥ 0 (i ∈ {1, . . . , n}) such that Pn := i=1 pi > 0.
Then one has the inequality
!
n
n
1 X
1 X
0≤
(6.81)
pi f (xi ) − f
pi xi
Pn i=1
Pn i=1
n
n
n
1 X
1 X
1 X
0
≤
pi xi f (xi ) −
pi xi ·
pi f 0 (xi ) .
Pn i=1
Pn i=1
Pn i=1
Proof. Since f is differentiable convex on I̊, one has
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0
(6.82)
f (x) − f (y) ≥ (x − y) f (y) ,
for any x, y ∈I̊.
If we choose x =
Pn
(6.83)
f
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1
Pn
i=1
n
1 X
pi xi
Pn i=1
pi xi and y = yk , k ∈ {1, . . . , n} , we get
!
− f (yk ) ≥
n
1 X
pi xi − yk
Pn i=1
!
0
f (yk ) .
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Multiplying (6.83) by pk ≥ 0 and summing over k from 1 to n, we deduce the
desired result (6.81).
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The following result holds [9].
Theorem 6.23. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a differentiable
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convex (concave) function on I̊ and abii ∈I̊ for each i ∈ {1, . . . , n} , and w̄ =
(w1 , . . . , wn ) is a sequence of nonnegative real numbers, then
!2
Pn
n
n
n
X
X
X
w
a
b
bi
i
i
i
2
2
2
wi ai
wi ai f
(6.84) 0 ≤
−
wi ai
· f Pi=1
n
2
a
i
i=1 wi ai
i=1
i=1
i=1
X
n
n
n
n
X
X
X
bi
bi
2
0
2 0
≤
wi ai
wi ai bi f
−
wi ai bi
wi ai f
.
a
a
i
i
i=1
i=1
i=1
i=1
Proof. Follows from Lemma 6.22 on choosing pi = wi a2i , xi =
bi
,i
ai
∈ {1, . . . , n} .
S.S. Dragomir
The following corollary holds [9].
Corollary 6.24. Let ā = (a1 , . . . , an ), b̄ = (b1 , . . . , bn ) be two sequences of
positive real numbers with ai 6= 0, i ∈ {1, . . . , n} . If p ∈ [1, ∞), then one has
the inequality
!2−p
!p
n
n
n
n
X
X
X
X
0≤
wi a2i
wi a2−p
bpi −
wi a2i
(6.85)
wi ai bi
i
i=1
"
≤p
i=1
n
X
wi a2i
i=1
n
X
wi a2−p
bi −
i
i=1
n
X
i=1
wi ai bi
i=1
i=1
n
X
#
wi a3−p
.
bp−1
i
i
i=1
0≤
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If p ∈ (0, 1) , then
(6.86)
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n
X
i=1
!2−p
wi a2i
n
X
i=1
!
wi ai bi
−
n
X
i=1
wi a2i
n
X
i=1
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bpi
i
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"
≤p
n
X
i=1
6.14.
wi ai bi
n
X
wi ai3−p bpi
i=1
−
n
X
wi a2i
i=1
n
X
#
wi a2−p
bi
i
.
i=1
An Inequality via a Refinement of Jensen’s Inequality
We will use the following lemma which contains a refinement of Jensen’s inequality obtained in [13].
Lemma 6.25. Let f : I ⊆PR →R be a convex function on the interval I and
xi ∈ I, pi ≥ 0 with Pn := ni=1 pi > 0. Then the following inequality holds:
!
n
n
X
xi1 + · · · + xik+1
1
1 X
pi · · · pik f
(6.87) f
pi xi ≤ k+1
Pn i=1
Pn i ,...,i =1 1
k+1
1
k+1
n
xi1 + · · · + xik
1 X
≤ k
pi · · · pik f
Pn i ,...,i =1 1
k
1
≤ ··· ≤
k
1
Pn
n
X
pi f (xi ) ,
i=1
where k ≥ 1, k ∈ N.
Proof. We shall follow the proof in [13].
The first inequality follows by Jensen’s inequality for multiple sums
x +···+x

 Pn
i1
ik+1
i1 ,...,ik+1 =1 pi1 · · · pik+1
k+1

Pn
f
p
·
·
·
p
i
i
k+1
i1 ,...,ik+1 =1 1
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x +···+x
i1
ik+1
p
·
·
·
p
f
i
i
k+1
i1 ,...,ik+1 =1 1
k+1
Pn
i1 ,...,ik+1 =1 pi1 · · · pik+1
Pn
=
since
x +···+x
i1
ik+1
n
p
·
·
·
p
X
ik+1
i1 ,...,ik+1 =1 i1
k+1
Pn
= Pnk
pi xi
i1 ,...,ik+1 =1 pi1 · · · pik+1
i=1
Pn
and
n
X
pi1 · · · pik+1 = Pnk+1 .
i1 ,...,ik+1 =1
Now, applying Jensen’s inequality for
xi + xi3 · · · + xik + xik+1
xi1 + xi2 · · · + xik−1 + xik
, y2 = 2
,
k
k
xi + xi1 + xi2 + · · · + xik−1
· · · , yk+1 = k+1
k
we have
y1 + y2 + · · · + yk + yk+1
f (y1 ) + f (y2 ) + · · · + f (yk ) + f (yk+1 )
f
≤
,
k+1
k+1
y1 =
which is equivalent to
xi1 + · · · + xik+1
(6.88) f
k+1
x +x ···+x +x x +x +x +···+x
i1
i2
ik−1
ik
ik+1
i1
i2
ik−1
f
+ ··· + f
k
k
.
≤
k+1
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Multiplying (6.88) with the nonnegative real numbers pi1 , . . . , pik+1 and summing over
i1 , . . . , ik+1 from 1 to n we deduce
(6.89)
n
X
pi1 · · · pik+1 f
i1 ,...,ik+1 =1
xi1 + · · · + xik+1
k+1

n
1  X
≤
k + 1 i ,...,i
1
+
= Pn
i1
n
X
pi1 · · · pik+1 f
k+1 =1
xi1 + · · · + xik
k
+ ···

xik+1 + xi1 + xi2 + · · · + xik−1

k
i1 ,...,ik+1 =1
n
X
xi1 + · · · + xik
pi1 · · · pik f
k
,...,i =1
pi1 · · · pik+1 f
k
which proves the second part of (6.87).
The following result holds.
Theorem 6.26. Let f : I ⊆ R →R be a convex function on the interval I, ā =
(a1 , . . . , an ), b̄ = (b1 , . . . , bn ) real numbers such that ai 6= 0, i ∈ {1, . . . , n}
and abii ∈ I, i ∈ {1, . . . , n} . If w̄ = (w1 , . . . , wn ) are positive real numbers,
then
Pn
wi ai bi
i=1
(6.90) f Pn
2
i=1 wi ai
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n
X
1
≤ Pn
k+1
( i=1 wi a2i )
 bi
i1 ,...,ik+1 =1
1
×f
wi1 · · · wik+1 a2i1 · · · a2ik+1
ai1
+ ··· +
k+1
bik+1

aik+1

b
i1
n
X
1
ai
≤ Pn
wi1 · · · wik a2i1 · · · a2ik f  1
k
( i=1 wi a2i ) i1 ,...,ik =1
n
X
1
bi
2
≤ · · · ≤ Pn
wi ai f
.
2
ai
i=1 wi ai i=1
+ ··· +
k
bik
aik


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The proof is obvious by Lemma 6.25 applied for pi = wi a2i , xi =
{1, . . . , n} .
The following corollary holds.
bi
,
ai
i ∈
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Corollary 6.27. Let ā, b̄ and w̄ be sequences of positive real numbers. If
p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequalities
!p
n
X
wi ai bi
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i=1
Pn
p−k−1
≤ ( i=1 wi a2i )
(≥)
(k + 1)p
n
X
i1 ,...,ik+1 =1
wi1 · · · wik+1 a2i1
· · · a2ik+1
bi
bi1
+ · · · + k+1
ai1
aik+1
p
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n
X
Pn
p−k
≤ ( i=1 wi a2i )
(≥)
kp
≤
≤
···
(≥)
(≥)
n
X
wi1 · · · wik a2i1
i1 ,...,ik =1
!p−1
wi a2i
i=1
n
X
· · · a2ik
bi1
bi
+ ··· + k
ai1
aik
p
wi ai2−p bpi .
i=1
Remark 6.4. If p = 2, then we deduce the following refinement of the (CBS) −
inequality

2
!2
P
n
n
k+1
k+1
1−k
n
X
X
( i=1 wi a2i )
X Y 
wi ai bi
≤
w
·
·
·
w
bi`
aij 

i
i
1
k+1
(k + 1)2
j=1
i=1
i1 ,...,ik+1 =1
`=1
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j6=`
P
2−k
( ni=1 wi a2i )
≤
k2
n
X

2
k
k
X Y 
bi`
aij 
wi1 · · · wik 
i1 ,...,ik =1
`=1
j=1
j6=`
n
1X
wi wj (bi aj + ai bj )2
≤ ··· ≤
4 i,j=1
≤
n
X
i=1
6.15.
wi a2i
n
X
wi b2i .
i=1
Another Refinement via Jensen’s Inequality
The following refinement of Jensen’s inequality holds (see [15]).
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Lemma 6.28. Let f : [a, b] → R be
Pan differentiable convex function on (a, b)
and xi ∈ (a, b) , pi ≥ 0 with Pn := i=1 pi > 0. Then one has the inequality
!
n
n
1 X
1 X
(6.91)
pi f (xi ) − f
pi xi
Pn i=1
Pn i=1
!
n
n
1 X
X
1
pi f (xi ) − f
pj xj ≥
Pn
P
n j=1
i=1
!
n
n
n
X
X
X
1
1
0 1
pj xj ·
pi xi −
pj xj ≥ 0.
− f
Pn j=1
Pn i=1
Pn j=1
Proof. Since f is differentiable convex on (a, b) , then for each x, y ∈ (a, b),
one has the inequality
0
f (x) − f (y) ≥ (x − y) f (y) .
(6.92)
(6.94) f (xi ) − f
−
n
1 X
xi −
pj xj
Pn j=1
!
f
0
n
1 X
pj xj
Pn j=1
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for each x, y ∈ (a, b) .
P
If we choose y = P1n nj=1 pj xj and x = xi , i ∈ {1, . . . , n} , then we have
!
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f (x) − f (y) − (x − y) f 0 (y) = |f (x) − f (y) − (x − y) f 0 (y)|
≥ ||f (x) − f (y)| − |x − y| |f 0 (y)||
n
1 X
pj xj
Pn j=1
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Using the properties of the modulus, we have
(6.93)
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≥ f (xi ) − f
n
1 X
pj xj
Pn j=1
! n
1 X
pj xj f 0
− xi −
Pn j=1
n
1 X
pj xj
Pn j=1
!
for any i ∈ {1, . . . , n} .
If we multiply (6.94) by pi ≥ 0, sum over i from 1 to n, and divide by
Pn > 0, we deduce
!
n
n
1 X
1 X
pi f (xi ) − f
pj xj
Pn i=1
Pn j=1
!
!
n
n
n
X
1 X
1 X
1
pi xi −
pj xj f 0
pj xj
−
Pn i=1
Pn j=1
Pn j=1
!
n
n
X
1
1 X
pi f (xi ) − f
pj xj ≥
Pn i=1 Pn j=1
!
n
n
1 X
0 1 X
− xi −
p j xj f
pj xj Pn j=1
Pn j=1
!
n
n
1 X
1 X
≥
pi f (xi ) − f
pj xj Pn
P
n j=1
i=1
!
n
n
n
1 X 1 X
0 1 X
−
pi xi −
pj xj · f
pj xj .
Pn
Pn
Pn
i=1
j=1
j=1
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Since
n
n
1 X
1 X
pi xi −
pj xj
Pn i=1
Pn j=1
!
= 0,
the inequality (6.91) is proved.
In particular, we have the following result for unweighted means.
Corollary 6.29. With the above assumptions for f and xi , one has the inequality
x1 + · · · + xn
f (x1 ) + · · · + f (xn )
(6.95)
−f
n
n
n 1 X
x
+
·
·
·
+
x
1
n
xi − f
≥
n
n
i=1
n n
X
0 x1 + · · · + xn 1 X
1
− f
·
xj ≥ 0.
xi −
n
n i=1 n j=1 The following refinement of the (CBS) −inequality holds.
Theorem 6.30. If ai , bi ∈ R, i ∈ {1, . . . , n} , then one has the inequality;
(6.96)
n
X
i=1
a2i
n
X
b2i
−
i=1
n
X
ai b i
n X
a2i
b2i
2
Pn 2 2
≥ Pn 2 Pn
b
a
b
i=1 i i=1
j=1 j j
j=1 bj
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1
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n
n
n
X
X
X
ai bi ≥ 0.
−2 ak b k ·
|bi | bj aj b j k=1
i=1
j=1
Proof. If we apply Lemma 6.28 for f (x) = x2 , we get
!2
n
n
1 X 2
1 X
(6.97)
pi xi −
pi xi
Pn i=1
Pn i=1
!2 n
n
X
X
1
1
≥ pi x2i −
pj xj Pn j=1
Pn i=1 n
n
n
1 X
1 X
1 X
pk xk ·
− 2
pi xi −
pj xj ≥ 0.
Pn
Pn
Pn j=1
i=1
k=1
If in (6.97), we choose pi = b2i , xi =
ai
,
bi
i ∈ {1, . . . , n} , we get
Pn 2
P
2
( ni=1 ai bi )
ai
i=1
(6.98) Pn 2 − Pn 2 2
(
b)
i=1 bi
i=1 i
!2 Pn
2
n
X
1
a
b
a
j j
Pj=1
≥ Pn 2
b2i · 2i −
n
2
b
b
b
i
i=1 i i=1
j=1 j
.
Pn
Pn b2 ai − Pn a b Pn b2 k=1 ak bk j=1 j j
j=1 j i=1 i bi
,
P
−2 P
·
n
n
2
2 b
b
i=1 i
i=1 i
which is clearly equivalent to (6.96).
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6.16.
An Inequality via Slater’s Result
◦
Suppose that I is an interval of real numbers with interior I and f : I → R
◦
is a convex function on I. Then f is continuous on I and has finite left and
◦
◦
right derivatives at each point of I . Moreover, if x, y ∈I and x < y, then
D− f (x) ≤ D+ f (x) ≤ D− f (y) ≤ D+ f (y) which shows that both D− f and
◦
D+ f are nondecreasing functions on I . It is also known that a convex function
must be differentiable except for at most countably many points.
For a convex function f : I → R, the subdifferential
◦ of f denoted by ∂f is
the set of all functions ϕ : I → [−∞, ∞] such that ϕ I ⊂ R and
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f (x) ≥ f (a) + (x − a) ϕ (a) for any x, a ∈ I.
(6.99)
It is also well known that if f is convex on I, then ∂f is nonempty, D+ f, D− f ∈
∂f and if ϕ ∈ ∂f, then
D− f (x) ≤ ϕ (x) ≤ D+ f (x)
(6.100)
◦
for every x ∈I . In particular, ϕ is a nondecreasing function.
◦
If f is differentiable convex on I , then ∂f = {f 0 } .
The following inequality is well known in the literature as Slater’s inequality
[16].
Lemma 6.31. Let f : I → R be a P
nondecreasing (nonincreasing)
convex funcPn
n
tion, xi ∈ I, pi ≥ 0 with Pn := i=1 pi > 0 and i=1 pi ϕ (xi ) 6= 0 where
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ϕ ∈ ∂f. Then one has the inequality:
Pn
n
pi xi ϕ (xi )
1 X
i=1
Pn
pi f (xi ) ≤ f
.
Pn i=1
i=1 pi ϕ (xi )
(6.101)
Proof. Firstly, observe that since, for example, f is nondecreasing, then ϕ (x) ≥
0 for any x ∈ I and thus
Pn
pi xi ϕ (xi )
Pi=1
(6.102)
∈ I,
n
i=1 pi ϕ (xi )
since it is a convex combination of xi with the positive weights
x ϕ (xi )
Pn i
, i = 1, . . . , n.
i=1 pi ϕ (xi )
A similar argument applies if f is nonincreasing.
Now, if we use the inequality (6.99), we deduce
(6.103)
f (x) − f (xi ) ≥ (x − xi ) ϕ (xi ) for any x, xi ∈ I, i = 1, . . . , n.
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Multiplying (6.103) by pi ≥ 0 and summing over i from 1 to n, we deduce
n
n
n
1 X
1 X
1 X
(6.104) f (x) −
pi f (xi ) ≥ x ·
pi ϕ (xi ) −
pi xi ϕ (xi )
Pn i=1
Pn i=1
Pn i=1
for any x ∈ I.
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If in (6.104) we choose
Pn
pi xi ϕ (xi )
x = Pi=1
,
n
i=1 pi ϕ (xi )
which, we have proved that it belongs to I, we deduce the desired inequality
(6.101).
If one would like to drop the assumption of monotonicity for the function f,
then one can state and prove in a similar way the following result.
Lemma
Pn 6.32. Let f : I → R be a convex function, xi ∈ I, pi ≥ 0 with Pn > 0
and i=1 pi ϕ (xi ) 6= 0, where ϕ ∈ ∂f. If
Pn
pi xi ϕ (xi )
Pi=1
(6.105)
∈ I,
n
i=1 pi ϕ (xi )
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then the inequality (6.101) holds.
The following result in connection to the (CBS) −inequality holds.
Theorem 6.33. Assume that f : R+ → R is a convex function
on R+ := [0, ∞),
Pn 2
bi
ai , bi ≥ 0 with ai 6= 0, i ∈ {1, . . . , n} and i=1 ai ϕ ai 6= 0 where ϕ ∈ ∂f.
(i) If f is monotonic nondecreasing (nonincreasing) in [0, ∞) then

 Pn
bi
X
n
n
a
b
ϕ
X
i
i
i=1
ai
bi
2
2


(6.106)
ai ϕ
≤
ai · f
Pn 2 bi .
a
i
aϕ
i=1
i=1
i=1
i
ai
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(ii) If
bi
a
b
ϕ
i=1 i i
ai
Pn 2 bi ≥ 0,
i=1 ai ϕ ai
Pn
(6.107)
then (6.106) also holds.
Remark 6.5. Consider the function f : [0, ∞) → R, f (x) = xp , p ≥ 1. Then f
is convex and monotonic nondecreasing and ϕ (x) = pxp−1 . Applying (6.106),
we may deduce the following inequality:
!p+1
!p
n
n
n
X
X
X
3−p p−1
2−p
p
(6.108)
p
ai b i
≤
a2i
ai b i
i=1
i=1
i=1
for p ≥ 1, ai , bi ≥ 0, i = 1, . . . , n.
6.17.
S.S. Dragomir
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An Inequality via an Andrica-Raşa Result
The following Jensen type inequality has been obtained in [17] by Andrica and
Raşa.
Lemma 6.34. Let f : [a, b] → R be a twice differentiable function and assume
that
m = inf f 00 (t) > −∞ and M = sup f 00 (t) < ∞.
t∈(a,b)
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P
If xi ∈ [a, b] and pi ≥ 0 (i = 1, . . . , n) with ni=1 pi = 1, then one has the
inequalities:

!2 
!
n
n
n
n
X
X
X
X
1 
(6.109) m
pi x2i −
pi xi  ≤
pi f (xi ) − f
pi xi
2
i=1
i=1
i=1
i=1

!2 
n
n
X
X
1
≤ M
pi x2i −
pi xi  .
2
i=1
i=1
Proof. Consider the auxiliary function fm (t) := f (t) − 12 mt2 . This function is
00
twice differentiable and fm
(t) ≥ 0, t ∈ (a, b) , showing that fm is convex.
Applying Jensen’s inequality for fm , i.e.,
!
n
n
X
X
pi fm (xi ) ≥ fm
pi xi ,
i=1
i=1
we deduce, by a simple calculation, the first inequality in (6.109).
The second inequality follows in a similar way for the auxiliary function
fM (t) = 21 M t2 − f (t) . We omit the details.
The above result may be naturally used to obtain the following inequality
related to the (CBS) −inequality.
Theorem 6.35. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) be two sequences of
real numbers with the property that there exists γ, Γ ∈ R such that
ai
(6.110)
−∞ ≤ γ ≤
≤ Γ ≤ ∞, for each i ∈ {1, . . . , n} ,
bi
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and bi 6= 0, i = 1, . . . , n. If f : (γ, Γ) → R is twice differentiable and
m = inf f 00 (t) and M = sup f 00 (t) ,
t∈(γ,Γ)
t∈(γ,Γ)
then we have the inequality

n
n
1 X 2 X 2
(6.111)
m
ai
bi −
2
i=1
i=1
≤
n
X
i=1
b2i
n
X
n
X
!2 
ai b i

i=1
ai
2
bi f
−
bi
 i=1
n
n
1 X 2 X 2
≤ M
ai
bi −
2
i=1
i=1
n
X
i=1
n
X
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!2 P
n
ai b i
2
i=1
bi
f Pn 2
i=1 bi

!2
ai b i
S.S. Dragomir
.
Title Page
i=1
b2
Proof. We may apply Lemma 6.34 for the choices pi = Pn i b2 and xi =
k=1 k
get
"P
Pn
2 #
n
2
a
a
b
1
i
i
m Pni=1 i2 − Pi=1
n
2
2
k=1 bk
k=1 bk
Pn 2 ai Pn
i=1 bi f
bi
ai b i
i=1
Pn 2
≤
− f Pn 2
k=1 bk
k=1 bk
"P
P
2 #
n
n
2
a
a
b
1
i i
≤ M Pni=1 i2 − Pi=1
n
2
2
b
b
k=1 k
k=1 k
ai
bi
to
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giving the desired result (6.111).
The following corollary is a natural consequence of the above theorem.
Corollary 6.36. Assume that ā, b̄ are sequences of nonnegative real numbers
and
ai
(6.112)
0<ϕ≤
≤ Φ < ∞ for each i ∈ {1, . . . , n} .
bi
Then for any p ∈ [1, ∞) one has the inequalities

!2 
n
n
n
X
X
X
1
(6.113)
p (p − 1) ϕp−2 
a2i
b2i −
ai bi 
2
i=1
i=1
i=1
!
!p
!2−p
p
n
n
n
n
X
X
X
X
p
2−p
≤
b2i
ai b i
−
b2i
ai b i
i=1
i=1
i=1

n
n
X
X
1
p−2 
2
≤ p (p − 1) Φ
ai
b2i −
2
i=1
i=1
i=1
n
X
!2 
ai b i

i=1
if p ∈ [2, ∞) and
(6.114)
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
!2 
n
n
n
X
X
X
1
p (p − 1) Φp−2 
a2i
b2i −
ai bi 
2
i=1
i=1
i=1
!
!2−p
!p
p
n
n
n
n
X
X
X
X
p 2−p
2
2
≤
bi
ai b i
−
bi
ai b i
i=1
i=1
i=1
i=1
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
n
n
X
X
1
p−2 
2
≤ p (p − 1) ϕ
ai
b2i −
2
i=1
i=1
n
X
!2 
ai b i

i=1
if p ∈ [1, 2] .
6.18.
An Inequality via Jensen’s Result for Double Sums
The following result for convex functions via Jensen’s inequality also holds
[18].
Lemma 6.37. Let f : R → R be a convex (concave) function and x̄ = (x1 , . . . , xn ) ,
p̄ =P
(p1 , . . . , pn ) real sequences with the property that pi ≥ 0 (i = 1, . . . , n)
and ni=1 pi = 1. Then one has the inequality:
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"P
(6.115) f
Pn
2
n
2
pi xi )
i=1 pi xi − (
Pn 2
Pi=1
2
n
i=1 i pi − (
i=1 ipi )
#
i
p
p
f
(∆x
·
∆x
)
k
l
1≤i<j≤n i j
,
Pn 2
Pn
2
i
p
−
(
ip
)
i
i
i=1
i=1
P
≤ (≥)
Contents
hP
j−1
k,l=i
where ∆xk := xk+1 − xk (k = 1, . . . , n − 1) is the forward difference.
Proof. We have, by Jensen’s inequality for multiple sums that
"P
#
Pn
2
n
2
p
x
−
(
p
x
)
i
i
i
i
(6.116)
f Pi=1
Pi=1
2
n
n
2p − (
i
i
i=1
i=1 ipi )
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"P
1≤i<j≤n
=f
pi pj (xi − xj )2
P
1≤i<j≤n
#
pi pj (j − i)2

2 (xj −xi )2
p
p
(j
−
i)
2
i
j
1≤i<j≤n
(j−i)

=f
P
2
p
p
(j
−
i)
1≤i<j≤n i j
P
(xj −xi )2
2
p
p
(j
−
i)
f
2
i
j
1≤i<j≤n
(j−i)
≤
=: I.
P
2
1≤i<j≤n pi pj (j − i)
P
On the other hand, for j > i, one has
(6.117)
xj − xi =
j−1
X
(xk+1 − xk ) =
j−1
X
k=i
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S.S. Dragomir
∆xk
k=i
Title Page
Contents
and thus
!2
j−1
(xj − xi )2 =
X
k=i
∆xk
j−1
=
X
∆xk · ∆xl .
k,l=i
Applying once more the Jensen inequality for multiple sums, we deduce
#
" Pj−1
#
"
(xj − xi )2
k,l=i ∆xk · ∆xl
(6.118)
=f
f
(j − i)2
(j − i)2
Pj−1
k,l=i f (∆xk · ∆xl )
≤ (≥)
(j − i)2
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and thus, by (6.118), we deduce
P
(6.119)
2
Pj−1
f (∆xk ·∆xl )
pi pj (j − i) k,l=i(j−i)2
I ≤ (≥)
P
2
1≤i<j≤n pi pj (j − i)
hP
i
P
j−1
·
∆x
)
p
p
f
(∆x
k
l
i
j
1≤i<j≤n
k,l=i
,
=
Pn 2
Pn
2
i=1 i pi − (
i=1 ipi )
1≤i<j≤n
and then, by (6.116) and (6.119), we deduce the desired inequality (6.115).
The following inequality connected with the (CBS) −inequality may be
stated.
Theorem 6.38. Let f : R → R be a convex (concave) function and ā = (a1 , . . . , an ) ,
b̄ = (b1 , . . . , bn ) , w̄ = (w1 , . . . , wn ) sequences of real numbers such that
bi 6= 0, wi ≥ 0 (i = 1, . . . , n) and not all wi are zero. Then one has the inequality
"P
(6.120) f
Pn
Pn
2
n
2
2
i=1 wi ai
i=1 wi bi − (
i=1 wi ai bi )
Pn
Pn 2 2
Pn
2
2
i=1 wi bi
i=1 i wi bi − (
i=1 iwi bi )
hP
Proof. Follows by Lemma 6.37 on choosing pi = wi b2i and xi =
1, . . . , n. We omit the details.
S.S. Dragomir
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i
j−1
ak
al
2 2
w
w
b
b
f
∆
·
∆
1≤i<j≤n i j i j
k,l=i
bk
bl
.
Pn
Pn 2 2
Pn
2
2
i=1 wi bi
i=1 i wi bi − (
i=1 iwi bi )
P
≤ (≥)
#
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,
bi
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6.19.
Some Inequalities for the Čebyšev Functional
For two sequences of real numbers ā = (a1 , . . . , aP
n ) , b̄ = (b1 , . . . , bn ) and
p̄ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and ni=1 pi = 1, consider the
Čebyšev functional
T p̄, ā, b̄ :=
(6.121)
n
X
p i ai b i −
i=1
n
X
i=1
p i ai
n
X
p i bi .
i=1
By Korkine’s identity [1, p. 242] one has the representation
(6.122)
n
1X
T p̄, ā, b̄ =
pi pj (ai − aj ) (bi − bj )
2 i,j=1
X
=
pi pj (aj − ai ) (bj − bi ) .
1≤i<j≤n
Using the (CBS) −inequality for double sums one may state the following result
2
(6.123)
T p̄, ā, b̄
≤ T (p̄, ā, ā) T p̄, b̄, b̄ ,
where, obviously
(6.124)
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n
1X
T (p̄, ā, ā) =
pi pj (ai − aj )2
2 i,j=1
X
=
pi pj (aj − ai )2 .
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1≤i<j≤n
The following result holds [14].
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Lemma 6.39. Assume that ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are real numbers
such that for each i, j ∈ {1, . . . , n} , i < j, one has
(6.125)
m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) ,
where m, M are given real numbers.
P
If p̄ = (p1 , . . . , pn ) is a nonnegative sequence with ni=1 pi = 1, then one
has the inequality
(6.126)
(m + M ) T p̄, ā, b̄ ≥ T (p̄, ā, ā) + mM T p̄, b̄, b̄ .
Proof. If we use the condition (6.125), we get
(6.127)
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S.S. Dragomir
[M (bi − bj ) − (ai − aj )] [(ai − aj ) − m (bi − bj )] ≥ 0
Title Page
for i, j ∈ {1, . . . , n} , i < j.
If we multiply in (6.127), then, obviously, for any i, j ∈ {1, . . . , n} , i < j
we have
(6.128)
(aj − ai )2 + mM (bj − bi )2 ≤ (m + M ) (aj − ai ) (bj − bi ) .
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Multiplying (6.128) by pi pj ≥ 0, i, j ∈ {1, . . . , n} , i < j, summing over i and
j, i < j from 1 to n and using the identities (6.122) and (6.124), we deduce the
required inequality (6.125).
The following result holds [14].
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Theorem 6.40. If ā, b̄, p̄ are as in Lemma 6.39 and M ≥ m > 0, then one has
the inequality providing a reverse for (6.123)
2
T p̄, ā, b̄
≥
(6.129)
4mM
2 T (p̄, ā, ā) T p̄, b̄, b̄ .
(m + M )
Proof. We use the following elementary inequality
αx2 +
(6.130)
1 2
y ≥ 2xy, x, y ≥ 0, α > 0
α
to get, for the choices
√
α=
1
1
mM > 0, x = T p̄, b̄, b̄ 2 ≥ 0, y = [T (p̄, ā, ā)] 2 ≥ 0
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S.S. Dragomir
the following inequality:
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√
Contents
(6.131)
1
mM T p̄, b̄, b̄ + √
T (p̄, ā, ā)
mM
1
1
≥ 2 T p̄, b̄, b̄ 2 [T (p̄, ā, ā)] 2 .
Using (6.130) and (6.131), we deduce
1
1
(m + M )
√
T p̄, ā, b̄ ≥ T p̄, b̄, b̄ 2 [T (p̄, ā, ā)] 2
2 mM
which is clearly equivalent to (6.129).
The following corollary also holds [14].
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Corollary 6.41. With the assumptions of Theorem 6.40, we have:
(6.132)
1
1
0 ≤ T p̄, b̄, b̄ 2 [T (p̄, ā, ā)] 2 − T p̄, ā, b̄
√
√ 2
M− m
√
≤
T p̄, ā, b̄
2 mM
and
(6.133)
2
0 ≤ T (p̄, ā, ā) T p̄, b̄, b̄ − T p̄, ā, b̄
2
(M − m)2 ≤
T p̄, ā, b̄ .
4mM
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S.S. Dragomir
The following result is useful in practical applications [14].
Theorem 6.42. Let f, g : [α, β] → R be continuous on [α, β] and differentiable
on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume
(6.134)
f 0 (x)
,
x∈(α,β) g 0 (x)
−∞ < γ = inf
f 0 (x)
= Γ < ∞.
0
x∈(α,β) g (x)
sup
If x̄ is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j and if we denote
by f (x̄) := (f (x1 ) , . . . , f (xn )) , then we have the inequality:
(γ + Γ) T (p̄, f (x̄) , g (x̄))
≥ T (p̄, f (x̄) , f (x̄)) + γΓT (p̄, g (x̄) , g (x̄))
P
for any p̄ with pi ≥ 0 (i ∈ {1, . . . , n}) , ni=1 pi = 1.
(6.135)
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Proof. Applying the Cauchy Mean-Value Theorem, there exists ξij ∈ (α, β)
(i < j) such that
(6.136)
f (xj ) − f (xi )
f 0 (ξij )
= 0
∈ [γ, Γ]
g (xj ) − g (xi )
g (ξij )
for i, j ∈ {1, . . . , n} , i < j. Then
f (xj ) − f (xi ) f (xj ) − f (xi )
− γ ≥ 0, 1 ≤ i < j ≤ n,
(6.137) Γ −
g (xj ) − g (xi )
g (xj ) − g (xi )
which, by a similar argument to the one in Lemma 6.39 will give the desired
result (6.135).
A Survey on
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S.S. Dragomir
The following corollary is natural [14].
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Corollary 6.43. With the assumptions in Theorem 6.42 and if Γ ≥ γ > 0, then
one has the inequalities:
(6.138)
[T (p̄, f (x̄) , g (x̄))]2
4γΓ
≥
T (p̄, f (x̄) , f (x̄)) T (p̄, g (x̄) , g (x̄)) ,
(γ + Γ)2
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1
2
1
2
(6.139) 0 ≤ [T (p̄, f (x̄) , f (x̄))] [T (p̄, g (x̄) , g (x̄))] − T (p̄, f (x̄) , g (x̄))
√
√ 2
Γ− γ
√
≤
T (p̄, f (x̄) , g (x̄))
2 γΓ
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and
(6.140)
0 ≤ T (p̄, f (x̄) , f (x̄)) T (p̄, g (x̄) , g (x̄)) − T 2 (p̄, f (x̄) , g (x̄))
(Γ − γ)2 2
≤
T (p̄, f (x̄) , g (x̄)) .
4γΓ
6.20.
Other Inequalities for the Čebyšev Functional
For two sequences of real numbers ā = (a1 , . . . , aP
n ) , b̄ = (b1 , . . . , bn ) and
p̄ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and ni=1 pi = 1, consider the
Čebyšev functional
(6.141)
T p̄, ā, b̄ =
n
X
p i ai b i −
i=1
n
X
p i ai
i=1
n
X
S.S. Dragomir
p i bi .
Title Page
i=1
By Sonin’s identity [1, p. 246] one has the representation
(6.142)
Contents
n
X
T p̄, ā, b̄ =
pi (ai − An (p̄, ā)) bi − An p̄, b̄ ,
i=1
where
An (p̄, ā) :=
n
X
j=1
p j aj ,
An p̄, b̄ :=
A Survey on
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Type Discrete Inequalities
n
X
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p j bj .
j=1
Using the (CBS) −inequality for weighted sums, we may state the following
result
2
(6.143)
T p̄, ā, b̄
≤ T (p̄, ā, ā) T p̄, b̄, b̄ ,
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where, obviously
(6.144)
T (p̄, ā, ā) =
n
X
pi (ai − An (p̄, ā))2 .
i=1
The following result holds [14].
Lemma 6.44. Assume that ā = (a1 , . . . , an ) , b̄ = (b1 P
, . . . , bn ) are real numn
bers, p̄ =
(p1 , . . . , pn ) are nonnegative numbers with i=1 pi = 1 and bi 6=
An p̄, b̄ for each i ∈ {1, . . . , n} . If
(6.145)
−∞ < l ≤
ai − An (p̄, ā)
≤ L < ∞ for all i ∈ {1, . . . , n} ,
bi − An p̄, b̄
where l, L are given real numbers, then one has the inequality
(6.146)
(l + L) T p̄, ā, b̄ ≥ T (p̄, ā, ā) + LlT p̄, b̄, b̄ .
(6.147)
L−
ai − An (p̄, ā)
bi − An p̄, b̄
ai − An (p̄, ā)
−l
bi − An p̄, b̄
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!
≥0
(ai − An (p̄, ā))2 + Ll bi − An p̄, b̄
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2
≤ (L + l) (ai − An (p̄, ā)) bi − An p̄, b̄
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I
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for each i ∈ {1, . . . , n} .
2
If we multiply (6.147) by bi − An p̄, b̄
≥ 0, we get
(6.148)
S.S. Dragomir
Title Page
Proof. Using (6.145) we have
!
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for each i ∈ {1, . . . , n} .
Finally, if we multiply (6.148) by pi ≥ 0, sum over i from 1 to n and use the
identity (6.142) and (6.144), we obtain (6.146).
Using Lemma 6.44 and a similar argument to that in the previous section,
we may state the following theorem [14].
Theorem 6.45. With the assumption of Lemma 6.44 and if L ≥ l > 0, then one
has the inequality
(6.149)
2
T p̄, ā, b̄
≥
4lL
2 T (p̄, ā, ā) T p̄, b̄, b̄ .
(L + l)
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S.S. Dragomir
The following corollary is natural [14].
Corollary 6.46. With the assumptions in Theorem 6.45 one has
Title Page
1
1 0 ≤ [T (p̄, ā, ā)] 2 T p̄, b̄, b̄ 2 − T p̄, ā, b̄
√
√ 2
L− l
√
≤
T p̄, ā, b̄ ,
2 lL
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(6.150)
and
(6.151)
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2
0 ≤ T (p̄, ā, ā) T p̄, b̄, b̄ − T p̄, ā, b̄
2
(L − l)2 ≤
T p̄, ā, b̄ .
4lL
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6.21.
Bounds for the Čebyšev Functional
The following result holds.
Theorem 6.47. Let ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) (with bi 6= bj for i 6= j)
be two sequences of real numbers with the property that there exists the real
constants m, M such that for any 1 ≤ i < j ≤ n one has
m≤
(6.152)
aj − ai
≤ M.
bj − bi
A Survey on
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Then we have the inequality
mT p̄, b̄, b̄ ≤ T p̄, ā, b̄ ≤ M T p̄, b̄, b̄ ,
P
for any nonnegative sequence p̄ = (p1 , . . . , pn ) with ni=1 pi = 1.
(6.153)
S.S. Dragomir
Title Page
2
Proof. From (6.152), by multiplying with (bj − bi ) > 0, one has
2
2
m (bj − bi ) ≤ (aj − ai ) (bj − bi ) ≤ M (bj − bi )
for any 1 ≤ i < j ≤ n, giving by multiplication with pi pj ≥ 0, that
X
X
m
pi pj (bi − bj )2 ≤
pi pj (aj − ai ) (bj − bi )
1≤i<j≤n
≤M
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1≤i<j≤n
X
Contents
2
pi pj (bi − bj ) .
1≤i<j≤n
Using Korkine’s identity (see for example Subsection 6.19), we deduce the desired result (6.153).
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The following corollary is natural.
Corollary 6.48. Assume that the sequence b̄ in Theorem 6.47 is strictly increasing and there exists m, M such that
(6.154)
m≤
∆ak
≤ M, k = 1, . . . , n − 1;
∆bk
where ∆ak := ak+1 − ak is the forward difference, then (6.153) holds true.
Proof. Follows from Theorem 6.47 on taking into account that for j > i and
from (6.154) one has
m
j−1
X
∆bk ≤
k=i
j−1
X
∆ak ≤ M
k=i
j−1
X
S.S. Dragomir
∆bk ,
k=i
giving m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) .
Another possibility is to use functions that generate similar inequalities.
Theorem 6.49. Let f, g : [α, β] → R be continuous on [α, β] and differentiable
on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume that
f 0 (x)
,
x∈(α,β) g 0 (x)
−∞ < m = inf
A Survey on
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Type Discrete Inequalities
f 0 (x)
= M < ∞.
0
x∈(α,β) g (x)
sup
If x̄ = (x1 , . . . , xn ) is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j
and if we denote f (x̄) := (f (x1 ) , . . . , f (xn )) , then we have the inequality
(6.155) mT (p̄, g (x̄) , g (x̄)) ≤ T (p̄, f (x̄) , g (x̄)) ≤ M T (p̄, g (x̄) , g (x̄)) .
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Proof. Applying the Cauchy Mean-Value Theorem, for any j > i there exists
ξij ∈ (α, β) such that
f (xj ) − f (xi )
f 0 (ξij )
= 0
∈ [m, M ] .
g (xj ) − g (xi )
g (ξij )
Then, by Theorem 6.47 applied for ai = f (xi ) , bi = g (xi ) , we deduce the
desired inequality (6.155).
The following inequality related to the (CBS) −inequality holds.
Theorem 6.50. Let ā, x̄, ȳ be sequences of real numbers such that xi 6= 0 and
y
yi
6= xjj for i 6= j, (i, j ∈ {1, . . . , n}) . If there exist the real numbers γ, Γ such
xi
that
aj − ai
≤ Γ for 1 ≤ i < j ≤ n,
(6.156)
γ ≤ yj
− xyii
xj
then we have the inequality

n
n
X
X
2
(6.157)
γ
xi
yi2 −
i=1
n
X
i=1
≤
x2i
n
X
ai xi yi −
i=1
 i=1
n
n
X
X
2

≤Γ
xi
yi2 −
i=1
i=1
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I
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i=1
n
X
S.S. Dragomir
JJ
J
!2 
xi yi
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n
X
ai x2i
i=1
n
X
i=1
n
X
xi yi
i=1
!2 
xi yi
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Proof. Follows by Theorem 6.47 on choosing pi =
and M = Γ. We omit the details.
x2
Pn i
2
k=1 xk
, bi =
yi
,
xi
m = γ
The following different approach may be considered as well.
Theorem 6.51. Assume that ā = (a1 , . . . , an ) , b̄ = (b1 , . . . , bn ) are sequences
of real
p̄ = (p1 , . . . , pn ) is a sequence
of nonnegative real numbers
Pnumbers,
Pn
n
with i=1 pi = 1 and bi 6= An p̄, b̄ := i=1 pi bi . If
(6.158)
−∞ < l ≤
ai − An (p̄, ā)
≤ L < ∞ for each i ∈ {1, . . . , n} ,
bi − An p̄, b̄
where l, L are given real numbers, then one has the inequality
(6.159)
lT p̄, b̄, b̄ ≤ T p̄, ā, b̄ ≤ LT p̄, b̄, b̄ .
2
Proof. From (6.158), by multiplying with bi − An p̄, b̄
> 0, we deduce
2
l bi − An p̄, b̄
≤ (ai − An (p̄, ā)) bi − An p̄, b̄
2
≤ L bi − An p̄, b̄ ,
for any i ∈ {1, . . . , n} .
By multiplying with pi ≥ 0, and summing over i from 1 to n, we deduce
l
n
X
i=1
pi bi − An p̄, b̄
2
≤
n
X
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.
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Using Sonin’s identity (see for example Section 6.20), we deduce the desired
result (6.159).
The following result in connection with the (CBS) −inequality may be stated.
Theorem 6.52. Let ā, x̄, b̄ be sequences of real numbers such that xi 6= 0 and
yi
1
y
2
6= Pn 2 An x ,
xi
x
i=1 xi
A Survey on
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for i ∈ {1, . . . , n} . If there exists the real numbers φ, Φ such that
ai − Pn1 x2 An x2 , a
i=1 i
≤ Φ,
(6.160)
φ≤
yi
Pn1 2 An x2 , y
−
xi
x
x
i=1
where x2 = (x21 , . . . , x2n ) and
(6.157).
y
x
=
S.S. Dragomir
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i
y1
, . . . , xynn
x1
, then one has the inequality
Proof. Follows by Theorem 6.51 on choosing pi =
L = Φ. We omit the details.
x2
Pn i
i=1
x2i
, bi =
yi
,
xi
l = φ and
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References
[1] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, 1993.
[2] A.M. OSTROWSKI, Vorlesungen über Differential und Integralrechnung
II, Birkhäuser, Basel, 1951.
[3] S.S. DRAGOMIR, Ostrowski’s inequality in complex inner product
spaces, submitted.
[4] S.S. DRAGOMIR, Another Ostrowski type inequality, submitted.
A Survey on
Cauchy-Bunyakovsky-Schwarz
Type Discrete Inequalities
[5] K. FAN AND J. TODD, A determinantal inequality, J. London Math. Soc.,
30 (1955), 58–64.
S.S. Dragomir
[6] S.S. DRAGOMIR AND V.M. BOLDEA, Some new remarks on CauchyBuniakowski-Schwartz’s inequality, Coll. Sci. Pap. Faculty of Science
Kragujevac, 21 (1999), 33–40.
Title Page
[7] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for
Real Numbers (Romanian), “Caiete Metodico-Ştiintifice”, No. 57, 1989,
pp. 24, Faculty of Mathematics, University of Timişoara, Romania.
[8] S.S. DRAGOMIR, D.M. MILOŠEVIĆ AND Š.Z. ARSLANAGIĆ, A
Cauchy-Buniakowski-Schwartz inequality for Lipschitzian functions,
Coll. Sci. Pap. of the Faculty of Science, Kragujevac, 14 (1991), 25–28.
[9] S.S. DRAGOMIR, Generalisation of the (CBS) −inequality via Jensen
type inequalities, submitted.
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[10] P. LAH AND M. RIBARIĆ, Converse of Jensen’s inequality for convex
functions, Univ. Beograd Publ. Elek. Fak. Ser. Math, Fiz., No. 412-460,
(1973), 201–205.
[11] J.B. DIAZ AND F.T. METCALF, Stronger forms of a class of inequalities
of G. Pólya-G. Szegö and L.V. Kantorovich, Bull. Amer. Math. Soc., 69
(1963), 415–419.
[12] S.S. DRAGOMIR AND N.M. IONESCU, Some converse of Jensen’s inequaliy and applications, Anal. Num. Theor. Approx., 23 (1994), 71–78.
[13] J.E. PEČARIĆ AND S.S. DRAGOMIR, A refinement of Jensen’s inequality and applications, Studia Univ. Babeş-Bolyai, Mathematica, 34(1)
(1989), 15–19.
[14] S.S. DRAGOMIR, Some inequalities for the Čebyšev functional, The Australian Math. Soc. Gazette., 29(3) (2002), 164–168.
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[15] S.S. DRAGOMIR AND F.P. SCARMOZZINO, A refinement of Jensen’s
discrete inequality for differentiable convex functions, Preprint, RGMIA
Res. Rep. Coll., 5(4) (2002), Article 12. [ONLINE: http://rgmia.
vu.edu.au/v5n4.html]
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[16] M.S. SLATER, A companion inequality to Jensen’s inequality, J. Approx.
Theory, 32 (1981), 160–166.
[17] D. ANDRICA AND I. RAŞA, The Jensen inequality: refinements and applications, Anal. Num. Theor. Approx., (Romania), 14 (1985), 155–167.
[18] S.S. DRAGOMIR, On an inequality for convex functions, submitted.
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Index
J−convex function, 34, 35
Čebyšev functional, 198, 269, 274
Čebyšev’s inequality, 21, 229
1-norm, 175
Cauchy-Schwarz, 91
Cauchy Mean-Value Theorem, 273
Cauchy-Bunyakovsky-Schwarz, 9
Cauchy-Schwarz, 91
CBS-inequality, 9, 24, 26, 29, 30, 32,
additive, 19, 128, 130, 145, 146, 149,
33, 35–37, 40, 52, 72, 90, 233,
150
244
Alzer, 91, 92, 171
complex numbers, 11
Andrica-Badea inequality, 184, 188,
complex sequences, 220
190
double sums, 234, 237
arithmetic mean, 74, 128, 135, 231
for Lipschitzian functions, 239
Arithmetic-Geometric inequality, 23
for weights, 68, 104
asynchronous, 21, 227, 228
functionals associated to, 136
sequence, 229
generalisation, 13, 24, 29, 31, 32,
35, 46
barycentric method, 144
generalisation
via Young’s inBinet-Cauchy’s identity, 12
equality, 26
inequality related to, 129
Callebaut’s inequality, 43, 89
real numbers, 9, 13, 48, 84, 107
Cassels’ inequality, 143, 150, 151
real sequence, 217
additive version, 145
refinement, 22, 65, 75, 81, 83, 84,
refinement, 188, 192
115, 118, 123, 127, 130, 132,
unweighted, 145, 146, 149, 193
254, 257
Cauchy, 9, 10
Cauchy-Bunyakovsky-Schwarz, 9
discrete, 86
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reverse, 151, 164, 165, 168, 171, identity, 12, 13, 85, 89, 120, 124, 172,
175, 178, 183, 186, 191, 198,
200, 276
202, 206
Binet-Cauchy, 12
type, 41
Cauchy-Binet, 236
weights, 21
Korkine’s, 269
Lagrange, 10, 13, 90, 120, 124,
concave function, 243, 244, 246, 247,
236
249
Sonin, 274
convex mapping, 233
index set mapping, 109, 110, 116, 124,
Daykin-Eliezer-Carlitz, 86, 88
125, 134
De Bruijn’s inequality, 83
inner product spaces, 10, 89, 121
Diaz-Metcalf inequality, 194, 197, 247
Jensen’s inequality, 232, 233, 250, 251
discriminant, 10, 48, 52
discrete, 243, 244
double sums, 234, 236, 237, 269
reverse, 244
Dragomir-Ionescu inequality, 247
refinement, 250, 254
Dunkl-Williams’ inequality, 89
reverse, 247
Euler’s indicator, 37, 238
Korkine’s identity, 269
Fan and Todd inequalities, 226, 227
Lagrange’s identity, 10, 13, 90, 120,
forward difference, 172, 176
124, 236
geometric mean, 74, 128, 135, 231
Lah-Ribarić inequality, 244
Grüss inequality, 188
Lipschitzian functions, 239
Greub-Rheinboldt inequality, 149, 150 Lupaş inequality, 188
harmonic mean, 74, 75, 231
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matrices, 234
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maximum, 144, 185, 186
refinement, 72
McLaughlin’s inequality, 84
Cassels’ inequality, 188, 192
Milne’s inequality, 88
CBS inequality, 22, 65, 75, 81, 83,
minimum, 145
84, 90, 115, 118, 123, 127,
monotonicity, 100, 104, 108, 110, 115,
130, 132, 254, 257
118, 171
due to Alzer and Zheng, 91
multiple sums, 250
due to Daykin-Eliezer-Carlitz, 86
in terms of moduli, 55, 63
n-tuple, 245
Jensen’s
inequality, 250, 254
non-proportional, 217, 219, 223
non-constant sequences, 76
Ostrowski’s inequality, 217, 219, 222
Pólya-Szegö inequality, 194
Schwitzer’s result, 188
p-norm, 180
sequence less than the weights, 67
Pólya-Szegö, 147, 148
Pólya-Szegö’s inequality, 147, 149,
sequence whose norm is 1, 59
194
via Dunkl-Williams’ Inequality,
parameter, 19
89
Popoviciu’s inequality, 37, 38
reverse inequality, 151, 164, 165, 168,
positive real numbers, 18, 22, 24, 63,
171, 175, 178, 180, 183, 186,
74, 75, 143, 147, 149, 155,
195, 198, 202, 206
191, 230, 232, 244, 246, 249,
complex numbers, 159
252, 253
Jensen, 244, 247
Power functions, 232
real numbers, 155
power series, 40
via Čebyšev Functional, 198
proportional, 9, 10, 14, 24
via Andrica-Badea, 184
quadratic polynomial, 9, 46
via Diaz-Metcalf, 194
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rotation, 83
Schweitzer inequality, 188
sequence, 74, 86, 101, 112, 155, 171,
174, 199, 200, 218
asynchronous, 227, 229
complex, 220
complex numbers, 11, 25, 49, 90,
151, 171, 176, 180, 194, 219,
223, 225
decreasing nonnegative, 171
natural numbers, 238
nonnegative, 18, 20, 128, 171,
227, 270
nonnegative numbers, 22
nonnegative real numbers, 25, 43,
44, 67, 104, 106, 143, 149,
171, 176, 180, 184, 191, 243,
246, 249
nonzero complex numbers, 90
nonzero real numbers, 35
positive, 165
positive and real numbers, 15
positive numbers, 14, 86
positive real numbers, 18, 22, 24,
63, 75, 143, 147, 149, 155,
191, 230, 232, 244, 246, 249,
253
real, 217, 227, 272
real numbers, 13, 16, 17, 19, 21,
36, 40, 41, 46, 55, 56, 58,
59, 63, 65, 67, 70, 72, 73, 75,
76, 82–84, 104, 106, 119, 169,
175, 179, 183, 184, 186, 188,
198, 202, 208, 217, 222, 226,
227, 234–236, 239–241, 243,
246, 248, 269, 274
synchronous, 229
set-additive, 134, 137
set-superadditive, 135, 137, 138, 140
set-superadditivity, 136
set-supermultiplicative, 135
Shisha-Mond inequalities, 164
Sonin’s identity, 274
strong monotonicity, 115, 118
strong superadditivity, 112, 113, 116
sup-norm, 171
superadditivity, 106, 107, 109, 112,
113, 116, 119, 123, 125
supermultiplicative, 128
supermultiplicity, 128, 134
synchronous, 229
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sequence, 229
triangle inequality, 13, 172
112, 113, 128
Young’s Inequality, 22, 24, 26
Wagner’s inequality, 46, 49
Zagier inequalities, 167, 168
weights, 21, 67, 68, 104, 106, 107, Zheng, 91
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