Volume 8 (2007), Issue 4, Article 102, 6 pp.
TWO NEW ALGEBRAIC INEQUALITIES WITH 2n VARIABLES
XIAO-GUANG CHU, CHENG-EN ZHANG, AND FENG QI
S UZHOU H ENGTIAN T RADING C O . LTD .
E ASTERN BAODAI ROAD , S UZHOU C ITY,
J IANGSU P ROVINCE , 215128, C HINA
srr345@163.com
D EPARTMENT OF I NFORMATION E NGINEERING ,
X UCHANG VOCATIONAL T ECHNICAL C OLLEGE ,
X UCHANG C ITY, H ENAN P ROVINCE ,
461000, C HINA
zhangchengen63@163.com
R ESEARCH I NSTITUTE OF M ATHEMATICAL I NEQUALITY T HEORY,
H ENAN P OLYTECHNIC U NIVERSITY,
J IAOZUO C ITY, H ENAN P ROVINCE ,
454010, C HINA
qifeng618@gmail.com
URL: http://rgmia.vu.edu.au/qi.html
Received 12 March, 2007; accepted 05 December, 2007
Communicated by D. Hinton
A BSTRACT. 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.
Key words and phrases: Algebraic inequality, Cauchy’s inequality, Combinatorial identity, Algebraic identity.
2000 Mathematics Subject Classification. 26D05; 26D07; 26D15.
1. M AIN 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
i=1
x2n−1
n
≥ 2n−2
.
P2n i
2n−1
2
(2n − 1)
k6=i (xi + xk )
Equality in (1.1) holds if and only if xi = xj for all 1 ≤ i, j ≤ 2n.
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.
078-07
2
X.-G. C HU , C H .-E. Z HANG ,
AND
F. Q I
Theorem 1.2. Let n ≥ 2 and yi for 1 ≤ i ≤ 2n be positive real numbers. Then
2n
X
(1.2)
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.
P
The notation i+n−2
y
in Theorem 1.2 could be illustrated with an example to clarify the
k=i
Pk|2n
12
meaning: If n = 5 then 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
2. T WO L EMMAS
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
(2.1)
(n − k)
= 4n−1 n.
k
k=0
Proof. It is well known that
n
n
n
n−1
=
, 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
− (2n − 1)
+
=n
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
=n
− 2n(2n − 1)
+ 2n(2n − 1)
k
k
−
1
k−2
k=0
k=1
k=2
n−1 n−2 n−3 X
X
X
2n − 1
2n − 2
2n
2
=n
− 2n(2n − 1)
+ 2n(2n − 1)
k
k
k
k=0
k=0
k=0
2n
n
22n−1 − 2 2n−1
4n−1 − 2n−2
−2
n−1
n−1
24 − n
=n
− 2n(2n − 1)
+ 2n(2n − 1)
2
2
2
2
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 102, 6 pp.
2n−2
n−2
http://jipam.vu.edu.au/
T WO N EW A LGEBRAIC I NEQUALITIES WITH 2n VARIABLES
n−1
=4
n+
2n−1
n
4n(2n − 1)
− n2
2n
n
− 2n(2n − 1)
2
2n−2
n−1
3
+2
2n−2
n−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
The proof of Lemma 2.1 is complete.
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
2n
n−1+i
X
X
yk|2n ,
(2.2)
An =
yi
i=1
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 ).
From this, it is obtained readily that
X
An =
yi yj −
n
X
yi yn+i
i=1
1≤i<j≤2n
by induction on n. Since
X
X
xi xj =
1≤i<j≤n
(yi + yi+n )(yj + yj+n ),
1≤i<j≤n
then
An =
X
n
X
yi yj −
yi yi+n =
i=1
1≤i<j≤2n
X
X
(yi + yi+n )(yj + yj+n ) =
1≤i<j≤n
xi xj ,
1≤i<j≤n
which means that identity (2.3) holds. The proof of Lemma 2.2 is complete.
3. P ROOFS OF THE M AIN R ESULTS
Proof of Theorem 1.1. By Cauchy’s inequality [1, 4], it follows that
(3.1)
2n
X
P2n
i=1
2n X
2n
X
x2n−1
i
k6=i (xi
+ xk
)2n−1
xi (xi + xj )2n−1 ≥
i=1 j6=i
2n
X
!2
xni
.
i=1
Consequently, it suffices to show
!2
2n
2n X
2n
X
X
(2n − 1)4n−1
xni
≥n
xi (xi + xj )2n−1
i=1
i=1 j6=i
⇐⇒ (2n − 1)4
n−1
2n
X
i=1
2n−1
x2n
i + (2n − 1)2
X
xni xnj
1≤i<j≤2n
2n 2n
n X
2n − 1
2n − 1 X X 2n−k k
≥n
+
xi
xj
k
2n
−
k
i=1 j6=i
k=0
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 102, 6 pp.
http://jipam.vu.edu.au/
4
X.-G. C HU , C H .-E. Z HANG ,
2n−2
⇐⇒ (2n − 1) 2
−n
2n
X
F. Q I
AND
x2n
i
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
≥n
+
xi
xj .
k
2n − k
i=1 j6=i
k=1
Since
n
k
(3.2)
X
2n
2n−1
(2n − 1)2
−n
xni xnj
n
1≤i<j≤2n
=
n
n−k
and
n
k
=
n−1
k
+
n−1
k−1
2n−2
+ (2n − 1) 2
, the above inequality becomes
−n
2n
X
x2n
i
−n
i=1
Utilization of
2n−2
2 2
P2n
2n
k=0
k
= 22n and
2n−1
− n + (2n − 1)2
2n
0
=
2n
2n
2n 2n
n−1 X
2n X X
k=1
k
xi2n−k xkj ≥ 0.
i=1 j6=i
= 1 yields
2n−1
X 2n
2n
−n
−n
n
k
k=1,k6=n
"
= 22n n − 2n − n
2n X
2n
k=0
k
#
− 2 = 0.
Substituting this into (3.2) gives
2n
X
( n−1 "
X
i=1,j=1,i6=j
q=0
(3.3)
22n−2 − n − n
q X
2n
k=1
2n−2q−2
#
k
xqi xqj
X
)
x2n−2q−k−2
xkj
i
k=0
× (xi − xj )2 ≥ 0,
Pq
2n
k
= 0 for q = 0. Employing (2.1) in the above inequality leads to
"
( n−1
#
)
p p n−1
X
X
X
X
2n
2n
(2n − 2p − 1) 22n−2 − n
= 22n−2 n2 − n
(2n − 2p − 1)
k
k
p=0
p=0
k=0
k=0
n−1
X
2n−2 2
2 2n
= 0.
=2
n −n
(n − k)
k
k=0
where
k=1
This implies that inequality (3.3) is equivalent to
(3.4)
#
2n
(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
2n
X
( n "
X
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 102, 6 pp.
k−1
X
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T WO N EW A LGEBRAIC I NEQUALITIES WITH 2n VARIABLES
2n
X
i=1,j=1,i6=j
( n−1 "
X k(k + 1)
2
k=1
k
X
p(p + 1) 2n
2n−2
−n
2
k−p
2
p=1
×xk−1
xk−1
i
j
5
#
2n−2k−2
X
)
x2n−p−4
xpj
i
(xi − xj )4 ≥ 0.
p=0
In order to prove (3.4), it is sufficient to show
(3.5)
n−1
X
(n − 1)[(n − 1) + 1] 2n−2
p(p + 1)
2n
2
−n
> 0.
2
2
n−p−1
p=1
Considering (2.1), it is sufficient to show
n−1
X
2n
(3.6)
(n − k)
> 22n−2 .
k
k=0
P
n
2n
2n
By virtue of nk = n−k
and 2n
k=0 k = 2 , inequality (3.6) can be rearranged as
n−1
X
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
(3.7)
(2n − 2k − 1)
+
>
.
(2k − 2n − 1)
k
k
n
k=0
k=n+1
2n
2n
+ n+1
> 2n
is equivalent to 2 > n+1
Since n ≥ 2 and n−1
, then inequalities (3.7), (3.6)
n
n
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 xi . From Cauchy’s
i=1 yi =
inequality [1, 4], it follows that
!2
2n
2n
X
X
yi2
An
≥
yi ,
Pi+n−2
y
y
i−1|2n
k|2n
k=i
i=1
i=1
where An is defined by (2.2) or (2.3) in Lemma 2.2. Therefore, it is sufficient to prove
!2
!2
2n
n
X
X
X
(n − 1)
yi
≥ 2nAn ⇐⇒ (n − 1)
xi
≥ 2n
xi xj
i=1
i=1
⇐⇒
(n − 1)
⇐⇒
X
n
X
1≤i<j≤n
x2i ≥ 2
i=1
X
xi xj
1≤i<j≤2n
(xi − xj )2 ≥ 0.
1≤i<j≤2n
The proof of Theorem 1.2 is complete.
Proof of Theorem 1.3. Let
(3.8)
2(n−1)
Bn = 2
k
X
2n
k(k + 1) − n
p(p + 1)
.
k−p
p=1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 102, 6 pp.
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6
X.-G. C HU , C H .-E. Z HANG ,
AND
F. Q I
Then
Bn+1
(3.9)
k
X
2n + 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
2n−2 2
, 4Bn +
k
X
p(p + 1)Ck−p
p=1
and
2n
2n
2n + 2
2n + 2
Cq − Cq+1 = 4n
−
− (n + 1)
−
q
q+1
q
q+1
2n + 2
2n + 2 − q
2n − q
2n
− (n + 1)
1−
= 4n
1−
q+1
q
q+1
q
2n 2q − 2n + 1
2n + 2 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
(3.10)
Cq > Cq+1 .
q+1
From the above inequality and the facts that
2(2n − 1)(n + 1) 2n
>0
(3.11)
Cn =
n+2
n
and 2n−q
> 0, it follows easily that Cq > 0. Consequently, we have Bn+1 > 4Bn , and then
q+1
Bk+2 > 4Bk+1 . As a result, utilization of (3.5) gives
Bk+1 > 0,
Bk+2 > 0,
Bk+3 > 0,
Bk+4 > 0,
The proof of inequality (1.3) is complete.
··· ,
Bk+(n−k) = Bn > 0.
R EFERENCES
[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)
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 102, 6 pp.
http://jipam.vu.edu.au/