TWO NEW ALGEBRAIC INEQUALITIES WITH 2n
VARIABLES
XIAO-GUANG CHU
CHENG-EN ZHANG
Suzhou Hengtian Trading Co. Ltd.
Eastern Baodai Road, Suzhou City,
Jiangsu Province, 215128, China
EMail: srr345@163.com
Department of Information Engineering,
Xuchang Vocational Technical College
1060 Xuchang City, Henan Province, 461000, China
EMail: zhangchengen63@163.com
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
Title Page
FENG QI
Research Institute of Mathematical Inequality Theory,
Henan Polytechnic University,
Jiaozuo City, Henan Province, 454010, China.
EMail: qifeng618@gmail.com
Received:
12 March, 2007
Accepted:
05 December, 2007
Communicated by:
D. Hinton
2000 AMS Sub. Class.:
26D05; 26D07; 26D15.
Key words:
Algebraic inequality, Cauchy’s inequality, Combinatorial identity, Algebraic
identity.
Abstract:
In this paper, by proving a combinatorial identity and an algebraic identity and by
using Cauchy’s inequality, two new algebraic inequalities involving 2n positive
variables are established.
Acknowledgements:
The authors would like to express heartily their thanks to the anonymous referees
for their valuable corrections and comments on the original version of this paper.
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1
Main results
3
2
Two Lemmas
4
3
Proofs of the Main Results
7
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
Title Page
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1.
Main results
When solving Question CIQ-103 in [2] and Question CIQ-142 in [5], the following
two algebraic inequalities involving 2n variables were posed.
Theorem 1.1. Let n ≥ 2 and xi for 1 ≤ i ≤ 2n be positive real numbers. Then
(1.1)
2n
X
x2n−1
n
i
≥ 2n−2
.
2n−1
2
(2n − 1)
k6=i (xi + xk )
P2n
i=1
Equality in (1.1) holds if and only if xi = xj for all 1 ≤ i, j ≤ 2n.
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
Theorem 1.2. Let n ≥ 2 and yi for 1 ≤ i ≤ 2n be positive real numbers. Then
(1.2)
2n
X
i=1
yi−1|2n
yi2
Pi+n−2
k=i
yk|2n
≥
2n
,
n−1
where m|2n means m mod 2n for all nonnegative integers m. Equality in (1.2)
holds if and only if yi = yj for all 1 ≤ i, j ≤ 2n.
Pi+n−2
The notation k=i
yk|2n in Theorem 1.2 could be illustrated with an example
P
to clarify the meaning: If n = 5 then 12
k=9 yk|10 = y9 + y10 + y1 + y2 .
In this article, by proving a combinatorial identity and an algebraic identity and by
using Cauchy’s inequality, these two algebraic inequalities (1.1) and (1.2) involving
2n positive variables are proved.
Moreover, as a by-product of Theorem 1.1, the following inequality is deduced.
Theorem 1.3. For n ≥ 2 and 1 ≤ k ≤ n − 1,
k
X
2n
22(n−1) k(k + 1)
(1.3)
<
p(p + 1)
.
k−p
n
p=1
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2.
Two Lemmas
In order to prove inequalities (1.1) and (1.2), the following two lemmas are necessary.
Lemma 2.1. Let n and k be natural numbers such that n > k. Then
n−1
X
2 2n
(n − k)
(2.1)
= 4n−1 n.
k
k=0
Proof. It is well known that
n
n
n−1
n
=
, k
=n
,
k
n−k
k
k−1
X
2n n
n−2
2n
k(k − 1)
= n(n − 1)
,
= 4n .
k
k−2
i
i=0
Then
n−1
X
2n
(n − k)
k
k=0
X
n−1 n−1 n−1
X
X
2n
2n
2n
2
=n
− (2n − 1)
k
+
k(k − 1)
k
k
k
k=0
k=0
k=0
n−1
n−1
n−1 X
X
X
2n
2n − 1
2n − 2
2
− 2n(2n − 1)
+ 2n(2n − 1)
=n
k
k−1
k−2
k=0
k=1
k=2
2
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
Title Page
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=n
2
n−1 X
2n
k=0
4n −
= n2
2
= 4n−1 n +
k
2n
n
− 2n(2n − 1)
n−2 X
2n − 1
k=0
2n−1
2
− 2n(2n − 1)
4n(2n − 1) 2n−1
− n2
n
k
−2
2
2n
n
+ 2n(2n − 1)
n−3 X
2n − 2
k=0
n−1
2n−1
4
n−1
+ 2n(2n − 1)
− 2n(2n − 1) 2n−2
+2
n−1
2
k
−
2n−2
n−1
2n−2
n−2
−2
2n−2
n−2
2
2n − 2
2
= 4 n + 2(2n − 1) − n(2n − 1) − n(2n − 1) − 2(2n − 1)(n − 1)
n−1
n−1
= 4 n.
n−1
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
The proof of Lemma 2.1 is complete.
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Lemma 2.2. Let n ≥ 2 and yi for 1 ≤ i ≤ 2n be positive numbers. Denote
xi = yi + yn+i for 1 ≤ i ≤ n and
Contents
(2.2)
An =
2n
X
i=1
yi
n−1+i
X
yk|2n ,
k=i+1
where m|2n means m mod 2n for all nonnegative integers m. Then
X
(2.3)
An =
xi xj .
1≤i<j≤n
Proof. Formula (2.2) can be written as
(2.4) An = y1 (y2 + · · · + yn ) + y2 (y3 + · · · + yn+1 ) + · · · + y2n (y1 + · · · + yn−1 ).
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From this, it is obtained readily that
X
An =
yi yj −
1≤i<j≤n
X
xi xj =
yi yn+i
i=1
1≤i<j≤2n
by induction on n. Since
X
n
X
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
(yi + yi+n )(yj + yj+n ),
1≤i<j≤n
vol. 8, iss. 4, art. 102, 2007
then
An =
X
1≤i<j≤2n
yi yj −
n
X
i=1
yi yi+n =
X
(yi + yi+n )(yj + yj+n ) =
1≤i<j≤n
X
xi xj ,
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1≤i<j≤n
which means that identity (2.3) holds. The proof of Lemma 2.2 is complete.
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3.
Proofs of the Main Results
Proof of Theorem 1.1. By Cauchy’s inequality [1, 4], it follows that
(3.1)
2n
X
i=1
2n X
2n
X
x2n−1
i
xi (xi + xj )2n−1 ≥
P2n
2n−1
k6=i (xi + xk )
i=1 j6=i
2n
X
!2
xni
.
i=1
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
Consequently, it suffices to show
!2
2n
2n X
2n
X
X
n−1
n
(2n − 1)4
xi
≥n
xi (xi + xj )2n−1
i=1
vol. 8, iss. 4, art. 102, 2007
i=1 j6=i
n−1
⇐⇒ (2n − 1)4
2n
X
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x2n
i
2n−1
+ (2n − 1)2
i=1
≥n
2n−2
⇐⇒ (2n − 1) 2
xni xnj
Contents
1≤i<j≤2n
n X
2n − 1
k=0
X
k
−n
2n
X
2n − 1
+
2n − k
X
2n X
2n
xi2n−k xkj
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i=1 j6=i
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x2n
i
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i=1
X
2n − 1
2n−1
+ (2n − 1)2
− 2n
xni xnj
n
1≤i<j≤2n
2n 2n
n−1
X 2n − 1
2n − 1 X X 2n−k k
xi
xj .
≥n
+
k
2n − k
i=1 j6=i
k=1
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Since
n
k
(3.2)
X
2n
2n−1
(2n − 1)2
−n
xni xnj
n
1≤i<j≤2n
n
n−k
=
n
k
and
=
2n−2
+ (2n − 1) 2
n−1
k
−n
n−1
k−1
+
2n
X
x2n
i
, the above inequality becomes
−n
i=1
Utilization of
2n−2
2 2
P2n
k=0
2n
k
= 22n and
− n + (2n − 1)2
2n−1
2n
0
2n 2n
n−1 X
2n X X
k=1
=
2n
2n
k
xi2n−k xkj ≥ 0.
i=1 j6=i
= 1 yields
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
2n−1
X 2n
2n
−n
−n
n
k
k=1,k6=n
" 2n #
X 2n
= 22n n − 2n − n
− 2 = 0.
k
k=0
Substituting this into (3.2) gives
)
( n−1 "
#
2n−2q−2
q 2n
X
X
X
X
2n
(3.3)
22n−2 − n − n
xqi xqj
x2n−2q−k−2
xkj
i
k
q=0
k=0
i=1,j=1,i6=j
k=1
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2
× (xi − xj ) ≥ 0,
where
Pq
k=1
2n
k
= 0 for q = 0. Employing (2.1) in the above inequality leads to
"
#
p n−1
X
X
2n
(2n − 2p − 1) 22n−2 − n
k
p=0
k=0
Close
= 22n−2 n2 − n
( n−1
X
(2n − 2p − 1)
p=0
2n−2 2
n −n
=2
n−1
X
p X
2n
k=0
2
(n − k)
k=0
2n
k
)
k
= 0.
This implies that inequality (3.3) is equivalent to
( n "
#
2n
k−1
X
X
X
2n
(3.4)
(xi − xj )2
k22n−2 − n
(k − q)
x2n−k−1
xk−1
i
j
q
q=0
i=1,j=1,i6=j
k=1
"
)
#
2n
2n
X
X
2n
+
(2n − k + 1)22n−2 − n
(2n − q + 1)
x2n−k
xk−2
≥ 0,
i
j
q−k
k=n+1
q=k
( n−1 "
X k(k + 1)
2n
X
i=1,j=1,i6=j
k=1
2
22n−2 − n
k
X
p=1
p(p + 1) 2n
2
k−p
×xk−1
xk−1
i
j
2n−2k−2
X
#
(xi − xj )4 ≥ 0.
p=0
In order to prove (3.4), it is sufficient to show
(3.5)
vol. 8, iss. 4, art. 102, 2007
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)
x2n−p−4
xpj
i
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
n−1
X
(n − 1)[(n − 1) + 1] 2n−2
p(p + 1)
2n
2
−n
> 0.
2
2
n
−
p
−
1
p=1
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Considering (2.1), it is sufficient to show
n−1
X
2n
(n − k)
> 22n−2 .
k
k=0
(3.6)
By virtue of
as
n
k
=
n
n−k
and
P2n
k=0
2n
k
= 22n , inequality (3.6) can be rearranged
n−1
X
(3.7)
2n
X
2n
2n
(n − k)
(k − n)
+
> 22n−1 ,
k
k
k=0
k=n+1
n−1
2n
X
X
2n
2n
2n
(2n − 2k − 1)
+
(2k − 2n − 1)
>
.
k
k
n
k=0
k=n+1
2n
n−1
2n
n+1
2n
n
n+1
,
n
Since n ≥ 2 and
then inequalities
+
>
is equivalent to 2 >
(3.7), (3.6) and (3.5) are valid. The proof of Theorem 1.1 is complete.
P
Pn
Proof of Theorem 1.2. By Lemma 2.2, it is easy to see that 2n
i=1 yi =
i=1 xi .
From Cauchy’s inequality [1, 4], it follows that
!2
2n
2n
2
X
X
yi
An
≥
yi ,
Pi+n−2
y
yk|2n
k=i
i=1 i−1|2n
i=1
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
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where An is defined by (2.2) or (2.3) in Lemma 2.2. Therefore, it is sufficient to
prove
(n − 1)
2n
X
!2
yi
≥ 2nAn ⇐⇒
n
X
(n − 1)
i=1
!2
≥ 2n
xi
i=1
⇐⇒
(n − 1)
n
X
X
xi xj
1≤i<j≤n
x2i ≥ 2
i=1
⇐⇒
X
X
xi xj
1≤i<j≤2n
(xi − xj )2 ≥ 0.
1≤i<j≤2n
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
The proof of Theorem 1.2 is complete.
Title Page
Proof of Theorem 1.3. Let
(3.8)
Contents
k
X
2n
2(n−1)
Bn = 2
k(k + 1) − n
p(p + 1)
.
k
−
p
p=1
Then
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Bn+1
(3.9)
JJ
k
X
2n + 2
2n−2 2
= k(k + 1)2
2 − (n + 1)
p(p + 1)
k−p
p=1
k
X
2n
2n + 2
= 4Bn +
p(p + 1) 4n
− (n + 1)
k−p
k−p
p=1
, 4Bn +
k
X
p=1
p(p + 1)Ck−p
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and
Cq − Cq+1
2n + 2
2n + 2
2n
2n
− (n + 1)
−
= 4n
−
q
q+1
q+1
q
2n
2n − q
2n + 2
2n + 2 − q
1−
= 4n
− (n + 1)
1−
q
q+1
q
q+1
2n + 2 2q − 2n − 1
2n 2q − 2n + 1
= 4n
− (n + 1)
q
q+1
q
q+1
2q − 2n + 1
>
Cq
q+1
for 0 ≤ q ≤ k − 1. Hence,
2n − q
Cq > Cq+1 .
q+1
Contents
From the above inequality and the facts that
Cn =
2(2n − 1)(n + 1) 2n
n+2
n
>0
Bk+2 > 0,
Bk+3 > 0,
Bk+4 > 0,
The proof of inequality (1.3) is complete.
··· ,
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and 2n−q
> 0, it follows easily that Cq > 0. Consequently, we have Bn+1 > 4Bn ,
q+1
and then Bk+2 > 4Bk+1 . As a result, utilization of (3.5) gives
Bk+1 > 0,
vol. 8, iss. 4, art. 102, 2007
Title Page
(3.10)
(3.11)
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
Bk+(n−k) = Bn > 0.
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References
[1] J.-CH. KUANG, Chángyòng Bùděngshì (Applied Inequalities), 3rd ed.,
Shāndōng Kēxué Jìshù Chūbǎn Shè (Shandong Science and Technology Press),
Jinan City, Shandong Province, China, 2004. (Chinese)
[2] J.-P. LI, Question CIQ-103, Communications in Studies of Inequalities, 11(2)
(2004), 277. (Chinese)
[3] ZH.-P. LIU, X.-G. CHU AND B.-N. GUO, On two algebraic inequalities with
even variables, J. Henan Polytech. Univ., 24(5) (2005), 410–414. (Chinese)
[4] D. S. MITRINOVIĆ, J. E. PEČARIĆ, AND A. M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, 1993.
[5] W.-J. ZHANG, Question CIQ-142, Communications in Studies of Inequalities,
12(1) (2005), 93. (Chinese)
Algebraic Inequalities with 2n
Variables
Xiao-Guang Chu, Cheng-En Zhang
and Feng Qi
vol. 8, iss. 4, art. 102, 2007
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