Journal of Inequalities in Pure and
Applied Mathematics
GENERALIZATIONS OF A CLASS OF INEQUALITIES
FOR PRODUCTS
volume 5, issue 3, article 77,
2004.
SHAN-HE WU AND HUAN-NAN SHI
Department of Mathematics
Longyan College
Longyan City, Fujian 364012, China.
EMail: wushanhe@yahoo.com.cn
Department of Electronic Information
Vocational-Technical Teacher’s College
Beijing Union University, Beijing 100011, China.
EMail: shihuannan@yahoo.com.cn
Received 20 January, 2004;
accepted 05 July, 2004.
Communicated by: F. Qi
Abstract
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c 2000 Victoria University
ISSN (electronic): 1443-5756
019-04
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Abstract
In this paper, a class of inequalities for products of positive numbers are generalized.
2000 Mathematics Subject Classification: Primary 26D15.
Key words: Inequality, Product, Arithmetic-geometric mean inequality, Jensen’s inequality.
The authors would like to express heartily many thanks to the anonymous referees
and to the Editor, Professor Dr. F. Qi, for their making great efforts to improve this
paper in language and mathematical expressions and typesetting.
Contents
1
Introduction and Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
Proofs of Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
References
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
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1.
Introduction and Main Results
In 1987, H.-Sh. Huang [2] proved the following algebraic inequality for products:
n
n Y
1
1
(1.1)
+ xi ≥ n +
,
xi
n
i=1
P
where x1 , x2 , . . . , xn are positive real numbers with ni=1 xi = 1.
In 2002, X.-Y. Yang [4] considered an analogous form of inequality (1.1)
and posed an interesting open problem as follows.
P
Open Problem. Assume x1 , x2 , . . . , xn are positive real numbers with ni=1 xi =
1 for n ≥ 3. Then
(1.2)
n Y
i=1
1
− xi
xi
≥
n−
1
n
n
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
Title Page
Contents
.
In [1], Ch.-H. Dai and B.-H. Liu gave an affirmative answer to the above
open problem.
In this article, by using the arithmetic-geometric mean inequality, inequalities (1.1) and (1.2) are refined and generalized as follows.
P
Theorem 1.1. Let x1 , x2 , . . . , xn be positive real numbers with ni=1 xi = k
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and k ≤ n, where k and n are natural numbers. Then we have for m ∈ N
2m
2m
! m(k2m −n2m )
m
n n
k
+n
m n Y
Y
1
n
k
nxi
m
+
x
≥
+
(1.3)
i
xm
k m nm
k
i
i=1
i=1
m
n
n
km
≥
+
.
k m nm
P
Theorem 1.2. Let x1 , x2 , . . . , xn be positive real numbers with ni=1 xi = k
for k ≤ 1 and n ≥ 3. Then for m ∈ N we have
(1.4)
! mn − m3
m
n n
m n Y
Y
1
n
k
nx
i
− xm
≥
−
i
m
x
k m nm
k
i
i=1
i=1
m
n
n
km
≥
−
.
k m nm
Remark 1.1. Choosing m = 1 and k = 1 in Theorem 1.1 and Theorem 1.2, we
can obtain inequalities (1.1) and (1.2) respectively.
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
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2.
Lemmas
To prove Theorem 1.1 and Theorem 1.2, we will use following lemmas.
P
Lemma 2.1. Let x1 , x2 , . . . , xn be positive real numbers with ni=1 xi = 1 and
n ≥ 3. Then
(2.1)
n Y
1
i=1
xi
− xi
≥
1
n−
n
n " Y
n
# n1 − 13
(nxi )
.
i=1
Generalizations of a Class of
Inequalities for Products
Proof. From the conditions of Lemma 2.1 and by using the arithmetic-geometric
mean inequality, we have for 1 ≤ p, q ≤ n and p 6= q
(2.2)
(1 − xp )(1 − xq ) = 1 − xp − xq + xp xq
X
=
xk + xp xq
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k6=p,q

=

X  xk
xk 
 + · · · +  + xp xq
n}
|n
{z
k6=p,q
n
1
# n(n−2)+1
Y x k n
≥ [n(n − 2) + 1]
xp xq
n
k6=p,q
" !n
# 1 2
n(n−2)
(n−1)
Y
1
2
= (n − 1)
xk xp xq
n
k6=p,q
"
Shan-He Wu and Huan-Nan Shi
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n(n−2)2
1 (n−1)
2
= (n − 1)
n
n
Y
!
n
(n−1)2
xi
1
(xp xq ) 1−n
k=1
then
(2.3)
n2 (n−2)2
1 2(n−1)
(1 − xi ) ≥ (n − 1)n
n
i=1
n
Y
n
Y
! n2 −2n+2
2
2(n−1)
xi
.
i=1
By the arithmetic-geometric mean inequality, we obtain


n
n
Y
Y1
1

(1 + xi ) =
 + · · · + +xi 
n
n
|
{z
}
i=1
i=1
n
)
(
1
n+1
n
Y
1 n
(2.4)
xi
≥
(n + 1)
n
i=1
1
! n+1
n2
n+1
n
Y
1
= (n + 1)n
xi
.
n
i=1
Utilizing (2.3) and (2.4) yields
n Y
1
(2.5)
− xi
xi
i=1
" n
#" n
# n
Y
Y
Y 1
=
(1 − xi )
(1 + xi )
x
i=1
i=1
i=1 i
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
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2
1
! n −2n+2
+ n+1
−1
n2
n2 (n−2)2 + n+1
n
2(n−1)2
Y
2(n−1)
1
n
n
≥ (n − 1) (n + 1)
xi
n
i=1
−n3 +3n2 −2n+2
"
#
n Y
n
2(n+1)(n−1)2
1
= n−
.
(nxi )
n
i=1
P
From the arithmetic-geometric mean inequality and ni=1 xi = 1 for n ≥ 3,
we have
!n
n
n
Y
X
(2.6)
0<
(nxi ) ≤
xi
= 1.
i=1
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−n3 + 3n2 − 2n + 2
1
= −
2
2(n + 1)(n − 1)
n
1
< −
n
≤ 0.
1 n(n − 3) (n2 + 2n + 8) + 10n + 6
−
3
6n(n + 1)(n − 1)2
1
3
(2.7)
i=1
# −n3 +3n2 −2n+2
2
"
2(n+1)(n−1)
(nxi )
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Therefore, by the monotonicity of the exponential function, we obtain
n
Y
Shan-He Wu and Huan-Nan Shi
i=1
Since n ≥ 3, it follows that
"
Generalizations of a Class of
Inequalities for Products
≥
n
Y
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# n1 − 13
(nxi )
Close
.
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i=1
J. Ineq. Pure and Appl. Math. 5(3) Art. 77, 2004
Combining inequalities (2.5) and (2.7) leads to inequality (2.1).
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P
Lemma 2.2. Let x1 , x2 , . . . , xn be positive real numbers with ni=1 xi = 1 for
n ≥ 3 and m a natural number. Then
# mn − m3
n " Y
n
n Y
1
1
(2.8)
− xm
≥ nm − m
(nxi )
.
i
m
x
n
i
i=1
i=1
Proof. Using the arithmetic-geometric mean inequality, we obtain


m−1
X
x2j
i
=
j=0
m−2
X
j=0

 x2j
x2j
i
i

 + x2m−2
+
·
·
·
+
i
 n2(m−j−1)
2(m−j−1) 
n
|
{z
}
Generalizations of a Class of
Inequalities for Products
n2(m−j−1)
(2.9)
≥
"m−1
X
j=0
#
n
2j
x2(m−1)
i
Pm−1
m−2
Y
j=0
n2m − 1
= 2(m−1) 2
(nxi )
n
(n − 1)
x2j
i
!n2(m−j−1) 
n2j

n2(m−j−1)
Pm−1
[2(m−j−1)]n2j
j=0
Pm−1 2j
n
j=0
j=0
(2.10)
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m−1
X 2j
1
− xi x1−m
xi
i
xi
j=0
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Pm−1
[2(m−j−1)]n2j
j=0
Pm−1 2j
n
j=0
1
n2m − 1
≥
− xi x1−m
(nxi )
i
xi
n2(m−1) (n2 − 1)
P
P
(m−1) m−1 n2j − m−1 2jn2j
j=0
j=1
2m
P
m−1
1
n −1
n2j
j=0
=
− xi
(nxi )
,
xi
nm−1 (n2 − 1)
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Hence
1
− xm
i =
xm
i
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and then
(2.11)
≥n
n Y
1
m
− xi
xm
i
i=1
n(1−m)
n2m − 1
n2 − 1
n "Y
n i=1
# (m−1)
# " Y
n
1
− xi
(nxi )
xi
i=1
Pm−1 2j Pm−1
n −
2jn2j
j=0
j=1
Pm−1 2j
n
j=0
In the following, we prove that for n ≥ 3
P
Pm−1
2j
2j
(m − 1) m−1
1 1
j=0 n −
j=1 2jn
(2.12)
≤ (m − 1)
−
.
Pm−1 2j
n 3
j=0 n
For m = 1, the equality in (2.12) holds. For m ≥ 2, we have
P
Pm−1
2j
2j
(m − 1) m−1
1 1
j=1 2jn
j=0 n −
(2.13)
− (m − 1)
−
Pm−1 2j
n 3
j=0 n
Pm−1 2j Pm−1
2j
(m − 1) 34 − n1
j=0 n −
j=1 2jn
=
Pm−1 2j
j=0 n
P
Pm−2
m−2 2j
2j
(m − 1) 43 − n1
j=0 n −
j=1 2jn
=
Pm−2 2j
j=0 n
(m − 1) n1 + 23 n2(m−1)
−
Pm−2 2j
j=0 n
.
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
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i P
2j
− n1 + 23 n2(m−1) − m−2
j=1 2jn
=
Pm−2 2j
j=0 n
1 4 1 2(m−1)
P
2j
(m − 1) 8 3 − n n
− n1 + 23 n2(m−1) − m−2
j=1 2jn
<
Pm−2 2j
j=0 n
P
9
2j
− 12 n2(m−1) − m−2
(m − 1) − 8n
j=1 2jn
=
Pm−2 2j
j=0 n
(m − 1)
h
n2(m−1) −1
n2 −1
4
3
−
1
n
Generalizations of a Class of
Inequalities for Products
< 0.
Hence inequality (2.12) holds.
Considering inequality (2.6) and the monotonicity of the exponential function and combining inequality (2.11) with (2.12) reveals
n Y
1
m
− xi
(2.14)
m
x
i
i=1
#(m−1)( n1 − 13 )
2m
n " Y
# " Y
n n
n −1
1
≥ nn(1−m)
− xi
(nxi )
.
2
n −1
xi
i=1
i=1
Substituting inequality (2.1) into (2.14) produces
2m
n n
n Y
1
n −1
1
m
n(1−m)
(2.15)
− xi ≥ n
n−
m
2−1
x
n
n
i
i=1
1
1
" n
#n−3 " n
#(m−1)( n1 − 13 )
Y
Y
×
(nxi )
(nxi )
i=1
i=1
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=
1
nm − m
n
n " Y
n
#m( n1 − 13 )
(nxi )
.
i=1
The proof is complete.
Lemma 2.3. Let x1 , x2 , . . . , xn be positive real numbers with
for n ≥ 3. Then for any natural number m, we have
(2.16)
Pn
i=1
xi = k ≤ 1
n Y
1
m
− xi
m
x
i
i=1
−n m
n Y
n m
n
km
k
xm
1
i
m
≥ n − m
−
− m .
n
k m nm
xm
k
i
i=1
Proof. It is easy to see that
Y
−1
n n m
n
Y
Y
1
k
xm
1 − x2m
i
i
m
nm
(2.17)
−
x
−
=
k
.
i
m
m
m
2m − x2m
x
x
k
k
i
i
i
i=1
i=1
i=1
f (x) = ln
1 − x2m
k 2m − x2m
for x ∈ (0, k), m ≥ 1 and k ≤ 1. Direct calculation shows that
(2.19)
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Define
(2.18)
Generalizations of a Class of
Inequalities for Products
f 0 (x) =
2m(1 − k 2m )x2m−1
,
(1 − x2m )(k 2m − x2m )
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(2.20)
2mx2(m−1) (1 − k 2m )
[(2m − 1)(1 − x2m )(k 2m − x2m )
(1 − x2m )2 (k 2m − x2m )2
+ 2mx2m (k 2m − x2m + 1 − x2m )]
≥ 0.
f 00 (x) =
This means that f is convex in the interval (0, k). Using Jensen’s inequality [3],
we obtain
P
n
1 − [ n1 ni=1 xi ]2m
1 − x2m
1X
i
(2.21)
ln
≥ ln
P
2m
n i=1 k 2m − x2m
k 2m − n1 ni=1 xi
i
P
for any 0 < xi < k ≤ 1 and i ∈ N. Using ni=1 xi = k in (2.21), it follows that
Generalizations of a Class of
Inequalities for Products
Shan-He Wu and Huan-Nan Shi
Title Page
n
Y
1 − x2m
i
≥
2m
k − x2m
i
i=1
(2.22)
k2m
n2m
2m
k 2m − nk 2m
1−
!n
Contents
,
therefore
(2.23)
k
nm
−n m
n
n
Y
1 − x2m
1
n
km
i
m
≥ n − m
−
.
k 2m − x2m
n
k m nm
i
i=1
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Substituting (2.17) into (2.23) leads to (2.16). The proof is complete.
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3.
Proofs of Theorems
Proof of Theorem 1.1. Using the arithmetic-geometric mean inequality, we obtain
1
1
1
xm
xm
i
i
m
+
x
=
+
·
·
·
+
+
+
·
·
·
+
i
2m xm
2m xm
2m
2m
xm
n
n
k
k
i
i
i
{z
}
|
{z
} |
2m
k
n2m
(3.1)
≥ (n2m + k 2m )
= (n
2m
+k
2m
"
) k
1
2m
n xm
i
−2mk2m
n2m xm
i
k 2m
1
k2m # k2m +n
2m
2m −mn2m
2m
n−2mn xmk
i
Generalizations of a Class of
Inequalities for Products
1
k2m +n2m
Shan-He Wu and Huan-Nan Shi
,
Title Page
therefore
(3.2)
Contents
n Y
i=1
1
+ xm
i
xm
i
≥ n2m + k
2m n
2m
2m
k −2mk n−2mn
n
k2m +n2m
n
Y
2m −n2m )
! m(k2m
2m
k
+n
xi
,
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that is
2m
(3.3)
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n Y
i=1
1
+ xm
i
m
xi
≥
2m
! m(k2m −n2m )
n Y
n
k
+n
n
k
nxi
+
.
k m nm
k
i=1
m
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From
that
Pn
i=1
xi = k and the arithmetic-geometric mean inequality, it follows
n
Y
nxi
(3.4)
i=1
k
≤
n
X
xi
i=1
!n
= 1,
k
and then, considering k ≤ n, we have
2m
(3.5)
2m
! m(k2m −n2m ) n
n Y
n
k
+n
n
k
nxi
nm k m
≥
+
.
+
k m nm
k m nm
k
i=1
m
m
Inequality (1.3) is then deduced by combining (3.3) and (3.5). This completes
the proof of Theorem 1.1.
P
Proof of Theorem 1.2. Applying ni=1 xki = 1 to Lemma 2.2, we have
(3.6)
n m
Y
k
i=1
xm
i
−
m
xm
k
i
≥
1
nm − m
n
! mn − m3
n Y
n
nxi
i=1
k
.
Substituting inequality (3.6) into Lemma 2.3 gives
−n m
n Y
n n m
Y
1
1
n
km
k
xm
i
m
m
− xi ≥ n − m
−
− m
xm
n
k m nm
xm
k
i
i
i=1
i=1
m
m
(3.7)
!n−3
m
n Y
n
n
km
nxi
≥
−
.
k m nm
k
i=1
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Since
0<
(3.8)
n
Y
nxi
i=1
and
m
n
−
m
3
≤
i=1
k
!n
=1
≤ 0, we have
(3.9)
k
n
X
xi
nm k m
−
k m nm
! mn − m3
n
m
n Y
n
km
nxi
n
.
≥
−
k m nm
k
i=1
Combining (3.7) and (3.9), we immediately obtain inequality (1.4). This completes the proof of Theorem 1.2.
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Inequalities for Products
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References
[1] Ch.-H. DAI AND B.-H. LIU, The proof of a conjecture, Shùxué Tōngxùn,
23 (2002), 27–28. (Chinese)
n
Q [2] H.-Sh. HUANG, On inequality ni=1 x1i + xi ≥ n + n1 , Shùxué
Tōngxùn, 5 (1987), 5–7. (Chinese)
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, 1993.
[4] X.-Y. YANG, The generalization of an inequality, Shùxué Tōngxùn, 19
(2002), 29–30. (Chinese)
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J. Ineq. Pure and Appl. Math. 5(3) Art. 77, 2004
http://jipam.vu.edu.au