Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 77, 2004
GENERALIZATIONS OF A CLASS OF INEQUALITIES FOR PRODUCTS
SHAN-HE WU AND HUAN-NAN SHI
D EPARTMENT OF M ATHEMATICS
L ONGYAN C OLLEGE
L ONGYAN C ITY, F UJIAN 364012, C HINA .
wushanhe@yahoo.com.cn
D EPARTMENT OF E LECTRONIC I NFORMATION
VOCATIONAL -T ECHNICAL T EACHER ’ S C OLLEGE
B EIJING U NION U NIVERSITY, B EIJING 100011, C HINA .
shihuannan@yahoo.com.cn
Received 20 January, 2004; accepted 05 July, 2004
Communicated by F. Qi
A BSTRACT. In this paper, a class of inequalities for products of positive numbers are generalized.
Key words and phrases: Inequality, Product, Arithmetic-geometric mean inequality, Jensen’s inequality.
2000 Mathematics Subject Classification. Primary 26D15.
1. I NTRODUCTION AND M AIN R ESULTS
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
n
n Y
1
1
(1.2)
− xi ≥ n −
.
xi
n
i=1
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
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.
019-04
2
S HAN -H E W U AND H UAN -NAN S HI
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 and k ≤ n, where
k and n are natural numbers. Then we have for m ∈ N
2m −n2m )
! m(k2m
m
n
m
n n
k
+n2m
m n Y
Y
1
n
k
n
km
nxi
m
(1.3)
≥
+
.
+ xi ≥
+
k m nm
xm
k m nm
k
i
i=1
i=1
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
! mn − m3
m
m
n
n n
m n Y
Y
1
n
n
k
nx
km
i
m
(1.4)
− xi ≥
−
≥
−
.
xm
k m nm
k
k m nm
i
i=1
i=1
Remark 1.3. Choosing m = 1 and k = 1 in Theorem 1.1 and Theorem 1.2, we can obtain
inequalities (1.1) and (1.2) respectively.
2. L EMMAS
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
# n1 − 13
n " Y
n n
Y
1
1
(2.1)
− xi ≥ n −
(nxi )
.
xi
n
i=1
i=1
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
(1 − xp )(1 − xq ) = 1 − xp − xq + xp xq
X
xk + xp xq
=
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
= (n − 1)2
xk xp xq
n
k6=p,q
! n 2
n(n−2)2 Y
n
(n−1)
(n−1)
1
1
(xp xq ) 1−n ,
= (n − 1)2
xi
n
k=1
"
(2.2)
then
(2.3)
n2 (n−2)2
1 2(n−1)
(1 − xi ) ≥ (n − 1)n
n
i=1
n
Y
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
n
Y
! n2 −2n+2
2
2(n−1)
xi
.
i=1
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G ENERALIZATIONS OF A C LASS OF I NEQUALITIES FOR P RODUCTS
3
By the arithmetic-geometric mean inequality, we obtain


n
n
Y
Y1
1

(1 + xi ) =
 + · · · + +xi 
|n {z n}
i=1
i=1
n
)
(
1
n+1
n
Y
1 n
(2.4)
≥
(n + 1)
xi
n
i=1
1
! n+1
n2
n+1
n
Y
1
.
xi
= (n + 1)n
n
i=1
Utilizing (2.3) and (2.4) yields
#" n
# n
"Y
n n
Y
Y
Y 1
1
(2.5)
− xi =
(1 − xi )
(1 + xi )
xi
x
i=1
i=1
i=1
i=1 i
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
xi
≥ (n − 1)n (n + 1)n
n
i=1
=
1
n−
n
n " Y
n
# −n3 +3n2 −2n+2
2
2(n+1)(n−1)
(nxi )
.
i=1
From the arithmetic-geometric mean inequality and
0<
(2.6)
n
Y
(nxi ) ≤
i=1
n
X
Pn
i=1
xi = 1 for n ≥ 3, we have
!n
xi
= 1.
i=1
Since n ≥ 3, it follows that
−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
Therefore, by the monotonicity of the exponential function, we obtain
"
(2.7)
n
Y
# −n3 +3n2 −2n+2
2
"
2(n+1)(n−1)
(nxi )
i=1
≥
n
Y
# n1 − 13
(nxi )
.
i=1
Combining inequalities (2.5) and (2.7) leads to inequality (2.1).
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
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
http://jipam.vu.edu.au/
4
S HAN -H E W U AND H UAN -NAN S HI
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
}
n2(m−j−1)
(2.9)
≥
Pm−1
#
"m−1
X
2(m−1)
n2j xi
j=0
=
m−2
Y
j=0
n
2m
−1
(nxi )
− 1)
x2j
i
n2(m−j−1)
!n2(m−j−1) 
Pm−1
[2(m−j−1)]n2j
j=0
Pm−1 2j
n
j=0
n2(m−1) (n2
j=0
n2j

.
Hence
1
− xm
i =
xm
i
m−1
X 2j
1
1−m
− xi xi
xi
xi
j=0
Pm−1
[2(m−j−1)]n2j
j=0
2m
Pm−1 2j
n
−
1
1
n
j=0
− 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
Pm−1 2j
1
n2m − 1
n
j=0
=
− xi
(nx
)
,
i
xi
nm−1 (n2 − 1)
(2.10)
and then
(2.11)
n Y
1
m
− xi
m
x
i
i=1
≥ nn(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
(m − 1)
(2.12)
Pm−1
j=0
n2j −
Pm−1
j=0
Pm−1
j=1
2jn2j
n2j
≤ (m − 1)
1 1
−
n 3
.
For m = 1, the equality in (2.12) holds. For m ≥ 2, we have
(2.13)
(m − 1)
Pm−1
j=0
n2j −
Pm−1
j=0
=
=
Pm−1
j=1
n2j
(m − 1)
4
3
−
1
n
2jn2j
− (m − 1)
Pm−1
j=0
n2j −
1 1
−
n 3
Pm−1
j=1
2jn2j
Pm−1
(m − 1)
4
3
2j
j=0 n
P
Pm−2
m−2 2j
2j
− n1
j=0 n −
j=1 2jn − (m − 1)
Pm−2 2j
j=0 n
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
1
n
+
2
3
n2(m−1)
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G ENERALIZATIONS OF A C LASS OF I NEQUALITIES FOR P RODUCTS
5
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
2(m−1) Pm−2
9
1
− j=1 2jn2j
(m − 1) − 8n − 2 n
=
Pm−2 2j
j=0 n
< 0.
(m − 1)
h
n2(m−1) −1
n2 −1
4
3
−
1
n
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
(2.14)
#(m−1)( n1 − 13 )
2m
n " Y
# " Y
n n n
Y
n
−
1
1
1
− xm
≥ nn(1−m)
− xi
(nxi )
.
i
m
2−1
x
n
x
i
i
i=1
i=1
i=1
Substituting inequality (2.1) into (2.14) produces
n Y
1
m
(2.15)
− xi
xm
i
i=1
# n1 − 13 " n
#(m−1)( n1 − 13 )
n n " Y
n
2m
Y
n
−
1
1
≥ nn(1−m)
n−
(nxi )
(nxi )
n2 − 1
n
i=1
i=1
1
1
#
m
−
n " Y
(
)
n
n
3
1
= nm − m
(nxi )
.
n
i=1
The proof is complete.
Pn
Lemma 2.3. Let x1 , x2 , . . . , xn be positive real numbers with i=1 xi = k ≤ 1 for n ≥ 3.
Then for any natural number m, we have
−n m
n Y
n n m
Y
1
1
n
km
k
xm
i
m
m
(2.16)
− xi ≥ n − m
− m
− m .
m
m
m
x
n
k
n
x
k
i
i
i=1
i=1
Proof. It is easy to see that
Y
−1
n n m
n
Y
Y
1
k
xm
1 − x2m
i
i
nm
m
(2.17)
−
x
−
=
k
.
i
m
m
m
2m
xi
xi
k
k − x2m
i
i=1
i=1
i=1
Define
f (x) = ln
(2.18)
1 − x2m
k 2m − x2m
for x ∈ (0, k), m ≥ 1 and k ≤ 1. Direct calculation shows that
(2.19)
f 0 (x) =
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
2m(1 − k 2m )x2m−1
,
(1 − x2m )(k 2m − x2m )
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6
S HAN -H E W U AND H UAN -NAN S HI
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) =
(2.20)
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
1X
1 − x2m
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
!n
n
k2m
Y
1
−
1 − x2m
i
n2m
(2.22)
≥
,
2m − x2m
k2m
2m
k
k
−
i
2m
n
i=1
therefore
k
(2.23)
nm
−n m
n
n
Y
1
n
km
1 − x2m
i
m
≥ n − m
− m
.
2m − x2m
m
k
n
k
n
i
i=1
Substituting (2.17) into (2.23) leads to (2.16). The proof is complete.
3. P ROOFS OF T HEOREMS
Proof of Theorem 1.1. Using the arithmetic-geometric mean inequality, we obtain
1
1
1
xm
xm
i
i
m
+
x
=
+
·
·
·
+
+
+
·
·
·
+
i
m
m
m
2m
2m
2m
2m
xi
n xi
n xi
k
{z k }
|
{z
} |
2m
k
n2m
(3.1)
≥ (n2m + k 2m )
"
1
2m
n xm
i
n2m xm
i
k 2m
1
k2m # k2m +n
2m
2m 1 2m
2m −mn2m
2m
2m
k
+n
= (n2m + k 2m ) k −2mk n−2mn xmk
,
i
therefore
(3.2)
n n
Y
−2mk2m −2mn2m k2m +n
2m
1
m
2m
2m n
+
x
≥
n
+
k
k
n
i
m
xi
i=1
n
Y
2m −n2m )
! m(k2m
2m
k
xi
+n
,
i=1
that is
m
n Y
n n
Y
1
n
km
nxi
m
+ xi ≥
+ m
m
m
xi
k
n
k
i=1
i=1
(3.3)
From
(3.4)
Pn
i=1
2m −n2m )
! m(k2m
2m
k
+n
.
xi = k and the arithmetic-geometric mean inequality, it follows that
!n
n
n
Y
X
nxi
xi
≤
= 1,
k
k
i=1
i=1
and then, considering k ≤ n, we have
2m
(3.5)
2m
! m(k2m −n2m ) n Y
n
n
k
+n
n
k
nxi
nm k m
+
≥
+
.
k m nm
k
k m nm
i=1
m
m
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
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G ENERALIZATIONS OF A C LASS OF I NEQUALITIES FOR P RODUCTS
7
Inequality (1.3) is then deduced by combining (3.3) and (3.5). This completes the proof of
Theorem 1.1.
Pn xi
Proof of Theorem 1.2. Applying i=1 k = 1 to Lemma 2.2, we have
! mn − m3
n Y
n
n m
Y
xm
1
nx
k
i
i
.
(3.6)
− m
≥ nm − m
m
x
k
n
k
i
i=1
i=1
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
− m
m
m
m
x
n
k
n
x
k
i
i
i=1
i=1
(3.7)
! mn − m3
m
n Y
n
n
km
nxi
≥
−
.
k m nm
k
i=1
Since
0<
(3.8)
n
Y
nxi
i=1
and
m
n
(3.9)
−
m
3
k
≤
n
X
xi
i=1
!n
k
=1
≤ 0, we have
! mn − m3
m
m
n
n
m n Y
n
k
nxi
n
km
−
≥
−
.
k m nm
k
k m nm
i=1
Combining (3.7) and (3.9), we immediately obtain inequality (1.4). This completes the proof
of Theorem 1.2.
R EFERENCES
[1] Ch.-H. DAI
nese)
AND
B.-H. LIU, The proof of a conjecture, Shùxué Tōngxùn, 23 (2002), 27–28. (Chi-
[2] H.-Sh. HUANG, On inequality
(Chinese)
Qn 1
i=1
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ
Kluwer Academic Publishers, 1993.
xi
AND
+ xi
≥
n+
1 n
n ,
Shùxué Tōngxùn, 5 (1987), 5–7.
A.M. FINK, Classical and New Inequalities in Analysis,
[4] X.-Y. YANG, The generalization of an inequality, Shùxué Tōngxùn, 19 (2002), 29–30. (Chinese)
J. Inequal. Pure and Appl. Math., 5(3) Art. 77, 2004
http://jipam.vu.edu.au/