Journal of Inequalities in Pure and
Applied Mathematics
INEQUALITIES ON LINEAR FUNCTIONS AND CIRCULAR POWERS
PANTELIMON STĂNICĂ
Auburn University Montgomery
Department of Mathematics
Montgomery, AL 36124-4023, USA.
EMail: stanica@strudel.aum.edu
URL: http://sciences.aum.edu/˜ stanpan
volume 3, issue 3, article 43,
2002.
Received 25 February, 2002;
accepted 10 April, 2002.
Communicated by: C.P. Niculescu
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
We prove some inequalities such as
x
x
F (x1 σ(1) , . . . , xnσ(n) ) ≤ F (xx1 1 , . . . , xxnn ),
where F is a linear function or a linear function in logarithms and σ is a permutation, which is a product of disjoint translation cycles. Stronger inequalities are
proved for second-order recurrence sequences, generalizing those of Diaz.
Inequalities on Linear Functions
and Circular Powers
Pantelimon Stănică
2000 Mathematics Subject Classification: 11B37, 11B39, 26D15.
Key words: Binary Sequences, Fibonacci Numbers, Linear Forms, Inequalities
The author would like to thank Professor C.P. Niculescu for his helpful comments,
which improved significantly the presentation.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
Further Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
References
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1.
Introduction
Define the second-order recurrent sequence by
xn+1 = axn + bxn−1 , x0 ≥ 0, x1 ≥ 1,
(1.1)
with a, b ≥ 1. If a = b = 1 and x0 = 0, x1 = 1 (or x0 = 2, x1 = 1),
then xn is the Fibonacci sequence, Fn (or Pell sequence, Pn ). Inequalities on
Fibonacci numbers were used recently by Bar-Noy et.al [1], to study a 9/8−
approximation for a variant of the problem that models the Broadcast Disks application (model for efficient caching of web pages). In [2], J.L. Diaz proposed
the following two inequalities:
F
F
(b)
<
Pantelimon Stănică
F
Fn
n+2
n+1
n+2
(a) FnFn+1 + Fn+1
+ Fn+2
< FnFn + Fn+1
+ Fn+2
,
Fn+2 Fn
FnFn+1 Fn+1
Fn+2
Inequalities on Linear Functions
and Circular Powers
Fn+1 Fn+2
FnFn Fn+1
Fn+2 .
In this note we show that the inequalities proposed by Diaz are not specific to
the Fibonacci sequence, holding for any strictly increasing sequence. Moreover,
we prove that stronger inequalities hold for any second-order recurrent sequence
as in (1.1). Furthermore, we pose a problem for future research.
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2.
The Results
We wondered if the inequalities (a), (b) were dependent on the Fibonacci sequence or if they can be extended to binary recurrent sequences. From here on,
we assume that all sequences have positive terms. Without too great a difficulty,
we prove, for a binary sequence, that
Theorem 2.1. For any positive integer n,
x
x
x
n+2
n+1
n+2
n
xxnn+1 + xn+1
+ xxn+2
< xxnn + xn+1
+ xn+2
,
xn+1 xn+2 xn
xn xn+1 xn+2
xn xn+1 xn+2 < xn xn+1 xn+2 .
(2.1)
(2.2)
Inequalities on Linear Functions
and Circular Powers
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Proof. We shall prove
xy + y ax+by + (ax + by)x < xx + y y + (ax + by)ax+by ,
(2.3)
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if 0 < x < y, which will imply our theorem. For easy writing, we denote
z = ax + by. Then (2.3) is equivalent to
(2.4)
xx xy−x − 1 + y y y z−y − 1 < z y z z−y − z −(y−x) .
x
y
y
y
y
y
Now, x + y < x + y < (x + y) ≤ z , since a, b ≥ 1. Moreover,
xy−x − 1 + y z−y − 1 = xy−x + y z−y − 2
< xz−y + y z−y − 1
< (x + y)z−y − 1
< z z−y − z −(y−x) .
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Taking A = xx , B = y y , C = xy−x − 1, D = y z−y − 1 and using the inequality
for positive numbers AC + BD ≤ (A + B)(C + D), we obtain (2.4).
The inequality (2.2) is implied by
xy y z z x < xx y y z z ⇐⇒ xy−x y z−y < z z−x .
(2.5)
But z z−x = z (z−y)+(y−x) ≥ (x + y)z−y (x + y)(y−x) > y z−y xy−x . The theorem
is proved.
Remark 2.1. We preferred to give this proof since it can be seen that the two
inequalities are far from being tight. We remark that inequality (2.2) can be
also shown by using Theorem 2.6.
Inequalities on Linear Functions
and Circular Powers
Pantelimon Stănică
With a little effort, while not attempting to have the best bound, we can
improve it, and also prove that the gaps are approaching infinity.
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Theorem 2.2. We have
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x
x
x
n+2
n+1
n+2
n
xxnn+1 + xn+1
+ xxn+2
< xxnn + xn+1
+ xn+2
− xn+2 xn+1 −xn
xn+2 xn
xn+1 xn+2
xn+1 xn
xn+2 < xxnn xn+1
xn+2 − 3xxnn xn+1
xn+2 .
xxnn+1 xn+1
In particular,
xn+1
xn+2 xn+2
n
lim [ xxnn + xn+1
+ xn+2
− xxnn+1 + xn+1
+ xxn+2
]=∞
n→∞
x
x
x
n+1
n+2
n+2 xn
lim [xxnn xn+1
xn+2
− xxnn+1 xn+1
xn+2 ] = ∞.
n→∞
In fact, the inequalities (2.1), (2.2) are not dependent on binary sequences,
at all. A much more general statement is true. Take σ a permutation, which
is a product of disjoint cyclic (translations by a fixed number, c(i) = i + t)
permutations.
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Theorem 2.3. Let n ≥ 2 and 1 ≤ x1 < x2 < . . . < xn a strictly increasing
sequence. Then
n
X
(2.6)
x
xi σ(i)
≤
i=1
n
X
xxi i ,
i=1
and
n
Y
(2.7)
x
xi σ(i)
≤
i=1
n
Y
xxi i ,
i=1
with strict inequality if σ is not the identity.
Inequalities on Linear Functions
and Circular Powers
Pantelimon Stănică
Proof. If σ is the identity permutation, the equality is obvious. Now, assume
that σ(i) = i + t. We take the case of t = 1 (the others are similar). We prove
(2.6) by induction on n. If n = 2, we need
xx1 2 + xx2 1 < xx1 1 + xx2 2 ,
which is equivalent to
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xx1 1 xx1 2 −x−1 − 1 < xx2 1 x2x2 −x−1 − 1 .
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The last inequality is certainly valid, since xx1 1 < xx2 1 and xx1 2 −x1 −1 < xx2 2 −x1 −
1.
Assuming the statement holds true for n, we prove it for n + 1. We need
Close
(2.8)
n+1
X
i=1
x
xi i+1
<
n+1
X
i=1
xxi i ,
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where xn+2 := x1 . We re-write (2.8) as
x
n+1
1
− xxn1 < xx1 1 + xx2 2 + · · · + xxnn + xn+1
,
(xx1 2 + xx2 3 + · · · + xxn1 ) + xxnn+1 + xxn+1
and using induction, it suffices to prove that
x
n+1
1
xxnn+1 + xxn+1
− xxn1 < xn+1
.
The previous inequality is equivalent to
xn+1 −x1
1
xn+1
−1 ,
xxn1 xxnn+1 −x1 − 1 < xxn+1
Inequalities on Linear Functions
and Circular Powers
which is obviously true, since xn < xn+1 .
The inequality (2.7) (when σ(i) = i + t) can be proved by induction, as well.
If n = 2, then
xx1 2 xx2 1 < xx1 1 xx2 2 ⇐⇒ xx1 2 −x1 < xx2 2 −x1 ,
which is true since x1 < x2 . Assuming the inequality holds true for n, we prove
it for n + 1. We need
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x
n+1
n
1
xx1 2 · · · xxnn+1 xxn1 = x1x2 · · · xxn−1
xxn1 xxnn+1 −x1 xxn+1
< xx1 1 · · · xn+1
.
Using the induction step, it suffices to prove
x
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x
n+1
n+1
1
xxnn+1 −x1 xxn+1
< xn+1
⇐⇒ xxnn+1 −x1 < xn+1
−x1
,
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which is valid since xn < xn+1 .
Now, take the general permutation σ 6=identity, which is a product of disjoint
cyclic permutations. Thus, σ can be written as a product of disjoint cycles as
J. Ineq. Pure and Appl. Math. 3(3) Art. 43, 2002
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σ = C1 × C2 × · · · × Cm . Recall that σ was taken such that all of its cycles
Ck are translations by a fixed number, say tk . Take Ck a cycle of length ek and
choose an index in Ck , say ik . Since σ is not the identity, then there is an index
k such that ek 6= 1. We write the inequalities (2.6) and (2.7) as
m eX
k −1
X
xσj+1 (i
k)
xσj (ik )
m eX
k −1
X
<
k=1 j=0
m eY
k −1
Y
xσj (i
)
xσj (ikk) ,
k=1 j=0
xσj+1 (i
k)
xσj (ik )
<
k=1 j=0
m eY
k −1
Y
xσj (i
)
xσj (ikk) ,
Inequalities on Linear Functions
and Circular Powers
k=1 j=0
Pantelimon Stănică
Therefore, it suffices to prove that, for any k, with ek 6= 1, we have
eX
k −1
xσj+1 (i
xσj (ik )
k)
<
j=0
eY
k −1
eX
k −1
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xσj (i
k)
xσj (ik ) ,
Contents
j=0
xσj+1 (i
xσj (ik )
k)
<
j=0
eY
k −1
xσj (i
k)
xσj (ik ) ,
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j=0
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that is,
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eX
k −1
xσj (i
xσj (ikk)
)+tk
<
eX
k −1
j=0
j=0
eY
k −1
eY
k −1
j=0
xσj (i
xσj (ikk)
)+tk
<
j=0
xσj (i
)
xσj (ikk) ,
xσj (i
)
xσj (ikk) .
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For k fixed, the above inequalities are just applications of the previous step (of
σ(i) = i + t), by taking a sequence yl to be xσj (ik ) in increasing order (we could
take from the beginning ik to be the minimum index in each cycle Ck ).
We can slightly extend the previous result (for a similar permutation σ) in
the following (we omit the proof).
Theorem 2.4. For any increasing sequence 0 < x1 < · · · < xn , we have
n
X
(2.9)
x
ai xi σ(i)
n
X
≤
i=1
i=1
n
Y
n
Y
x
ai xi σ(i)
≤
i=1
ai xxi i , and
Inequalities on Linear Functions
and Circular Powers
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ai xxi i ,
i=1
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where ai ≥ 0.
Contents
A parallel result involving logarithms is also true (σ is a permutation as before).
Theorem 2.5. For any finite increasing sequence, 0 < x1 < x2 < · · · < xn ,
and any positive real numbers ai , we have
n
X
ai xσ(i) log(xi ) ≤
n
X
i=1
i=1
n
Y
n
Y
i=1
ai xσ(i) log(xi ) =
i=1
ai xi log(xi ), and
ai xi log(xi ).
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The second identity is easily true since every ai , xi and log xi occurs in both
sides. We omit the proof of the first inequality, since it can be deduced easily
(as the referee observed) from the known fact (see [3, p. 261])
Theorem 2.6. Given two increasing sequences u1 ≤ u2 ≤ · · · ≤ un and
w1 ≤ w2 ≤ · · · ≤ wn , then
n
X
ui wn+1−i ≤
i=1
n
X
i=1
uτ (i) wσ(i) ≤
n
X
i=1
ui wi ,
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and Circular Powers
for any permutations σ, τ .
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3.
Further Comments
We believe that other inequalities of the type occurring in our theorems can also
be constructed. Let F : Rn → R be a function, with the properties
(3.1)
If xi ≤ yi , i = 1, . . . , n, then F (x1 , . . . , xn ) ≤ F (y1 , . . . , yn ),
with strict inequality if there is an index i such that xi < yi .
and
(3.2)
For 0 < x1 < x2 < · · · < xn , then,
F (xx1 2 , xx2 3 , . . . , xxn1 ) ≤ F (xx1 1 , xx2 2 , . . . , xxnn ).
Pn
As examples, we have the linear polynomial
PnF (x1 , . . . , xn ) = i=1 ai xi , the
linear form in logarithms F (x1 , . . . , xn ) = i=1 ai log(xi ), and the corresponding products.
We ask for more examples of functions satisfying (3.1) and (3.2), which cannot be derived trivially from the previous examples (by raising each variable to
the same power, for instance). Is it true that any symmetric polynomial satisfies
(3.1) and (3.2)? In addition to more examples, it might be worth investigating
the general form of polynomial functions that satisfy these properties.
This looks like a mathematical version of the philosophy saying:
Going one step at the time it is far better than jumping too fast and then at the
end falling to the bottom.
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and Circular Powers
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References
[1] A. BAR-NOY, R. BHATIA, J. NAOR AND B. SCHIEBER, Minimizing Service and Operating Costs of Periodic Scheduling, Symposium on Discrete
Algorithms (SODA), p. 36, 1998.
[2] J.L. DIAZ, Problem H-581, Fibonacci Quart., 40(1) (2002).
[3] G. HARDY, J.E. LITTLEWOOD, G. PÒLYA, Inequalities, Cambridge
Univ. Press, 2001.
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and Circular Powers
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