TWO NEW ALGEBRAIC INEQUALITIES WITH 2n VARIABLES

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Algebraic Inequalities with2n Variables

Xiao-Guang Chu, Cheng-En Zhang and Feng Qi

vol. 8, iss. 4, art. 102, 2007

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TWO NEW ALGEBRAIC INEQUALITIES WITH 2n VARIABLES

XIAO-GUANG CHU CHENG-EN ZHANG

Suzhou Hengtian Trading Co. Ltd. Department of Information Engineering, Eastern Baodai Road, Suzhou City, Xuchang Vocational Technical College

Jiangsu Province, 215128, China 1060 Xuchang City, Henan Province, 461000, China

EMail:srr345@163.com EMail:zhangchengen63@163.com

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 involving2npositive 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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vol. 8, iss. 4, art. 102, 2007

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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 involving2nvariables were posed.

Theorem 1.1. Letn ≥2andx_i for1≤i≤2nbe positive real numbers. Then

(1.1)

2n

X

i=1

x²ⁿ⁻¹_i P2n

k6=i(x_i +x_k)²ⁿ⁻¹ ≥ n

2²ⁿ⁻²(2n−1). Equality in (1.1) holds if and only ifx_i =x_j for all1≤i, j ≤2n.

Theorem 1.2. Letn ≥2andy_i for1≤i≤2nbe positive real numbers. Then

(1.2)

2n

X

i=1

y_i² yi−1|2nPi+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 ify_i =y_j for all1≤i, j ≤2n.

The notationPi+n−2

k=i y_k|2nin Theorem 1.2could be illustrated with an example to clarify the meaning: Ifn= 5thenP12

k=9yk|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 2npositive variables are proved.

Moreover, as a by-product of Theorem1.1, the following inequality is deduced.

Theorem 1.3. Forn≥2and1≤k ≤n−1,

(1.3)

k

X

p=1

p(p+ 1) 2n

k−p

< 2²⁽ⁿ⁻¹⁾k(k+ 1)

n .

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2. Two Lemmas

In order to prove inequalities (1.1) and (1.2), the following two lemmas are neces- sary.

Lemma 2.1. Letnandkbe natural numbers such thatn > k. Then (2.1)

n−1

X

k=0

(n−k)² 2n

k

= 4ⁿ⁻¹n.

Proof. It is well known that

n k

= n

n−k

, k n

k

=n

n−1 k−1

,

k(k−1) n

k

=n(n−1)

n−2 k−2

,

2n

X

i=0

2n i

= 4ⁿ.

Then

n−1

X

k=0

(n−k)² 2n

k

=n²

n−1

X

k=0

2n k

−(2n−1)

n−1

X

k=0

k 2n

k

+

n−1

X

k=0

k(k−1) 2n

k

=n²

n−1

X

k=0

2n k

−2n(2n−1)

n−1

X

k=1

2n−1 k−1

+ 2n(2n−1)

n−1

X

k=2

2n−2 k−2

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=n²

n−1

X

k=0

2n k

−2n(2n−1)

n−2

X

k=0

2n−1 k

+ 2n(2n−1)

n−3

X

k=0

2n−2 k

=n²4ⁿ− ²ⁿ_n

2 −2n(2n−1)2²ⁿ⁻¹−2 ²ⁿ⁻¹_n−1

2 + 2n(2n−1)4ⁿ⁻¹− ²ⁿ⁻²_n−1

−2 ²ⁿ⁻²_n−2 2

= 4ⁿ⁻¹n+ 4n(2n−1) ²ⁿ⁻¹_n

−n^{2 2n}_n

−2n(2n−1) 2n−2 n−1

+ 2 ²ⁿ⁻²_n−2 2

= 4ⁿ⁻¹n+

2(2n−1)²−n(2n−1)−n(2n−1)−2(2n−1)(n−1)

2n−2 n−1

= 4ⁿ⁻¹n.

The proof of Lemma2.1is complete.

Lemma 2.2. Let n ≥ 2 and y_i for 1 ≤ i ≤ 2n be positive numbers. Denote x_i =y_i+y_n+ifor1≤i≤nand

(2.2) A_n =

2n

X

i=1

y_i

n−1+i

X

k=i+1

y_k|2n,

wherem|2nmeansm mod 2nfor all nonnegative integersm. Then

(2.3) A_n = X

1≤i<j≤n

x_ix_j.

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 A_n= X

1≤i<j≤2n

y_iy_j −

n

X

i=1

y_iy_n+i

by induction onn. Since X

1≤i<j≤n

x_ix_j = X

1≤i<j≤n

(y_i+y_i+n)(y_j+y_j+n),

then

A_n = X

1≤i<j≤2n

y_iy_j−

n

X

i=1

y_iy_i+n = X

1≤i<j≤n

(y_i+y_i+n)(y_j+y_j+n) = X

1≤i<j≤n

x_ix_j,

which means that identity (2.3) holds. The proof of Lemma2.2is complete.

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3. Proofs of the Main Results

Proof of Theorem1.1. By Cauchy’s inequality [1,4], it follows that

(3.1)

2n

X

i=1

x²ⁿ⁻¹_i P2n

k6=i(xi+xk)²ⁿ⁻¹

2n

X

i=1 2n

X

j6=i

x_i(x_i+x_j)²ⁿ⁻¹ ≥

2n

X

i=1

xⁿ_i

!2

.

Consequently, it suffices to show (2n−1)4ⁿ⁻¹

2n

X

i=1

xⁿ_i

!2

≥n

2n

X

i=1 2n

X

j6=i

x_i(x_i+x_j)²ⁿ⁻¹

⇐⇒(2n−1)4ⁿ⁻¹

2n

X

i=1

x²ⁿ_i + (2n−1)2²ⁿ⁻¹ X

1≤i<j≤2n

xⁿ_ixⁿ_j

≥n

n

X

k=0

2n−1 k

+

2n−1 2n−k

²ⁿ X

i=1 2n

X

j6=i

x^2n−k_i x^k_j

⇐⇒(2n−1) 2²ⁿ⁻²−n

2n

X

i=1

x²ⁿ_i

+

(2n−1)2²ⁿ⁻¹−2n

2n−1 n

X

1≤i<j≤2n

xⁿ_ixⁿ_j

≥n

n−1

X

k=1

2n−1 k

+

2n−1 2n−k

²ⁿ X

i=1 2n

X

j6=i

x^2n−k_i x^k_j.

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Since ⁿ_k

= _n−kⁿ

and ⁿ_k

= ⁿ⁻¹_k

+ ⁿ⁻¹_k−1

, the above inequality becomes (3.2)

(2n−1)2²ⁿ⁻¹−n 2n

n

X

1≤i<j≤2n

xⁿ_ixⁿ_j

+ (2n−1) 2²ⁿ⁻²−n

2n

X

i=1

x²ⁿ_i −n

n−1

X

k=1

2n k

²ⁿ X

i=1 2n

X

j6=i

x^2n−k_i x^k_j ≥0.

Utilization ofP2n k=0

2n k

= 2²ⁿand ²ⁿ₀

= ²ⁿ_2n

= 1yields

2 2²ⁿ⁻² −n

+ (2n−1)2²ⁿ⁻¹−n 2n

n

−n

2n−1

X

k=1,k6=n

2n k

= 2²ⁿn−2n−n

" _2n X

k=0

2n k

−2

#

= 0.

Substituting this into (3.2) gives

(3.3)

2n

X

i=1,j=1,i6=j

(_n−1 X

q=0

"

2²ⁿ⁻²−n−n

q

X

k=1

2n k

# x^q_ix^q_j

2n−2q−2

X

k=0

x^{2n−2q−k−2}_i x^k_j )

×(x_i−x_j)² ≥0, wherePq

k=1 2n

k

= 0forq= 0. Employing (2.1) in the above inequality leads to

n−1

X

p=0

(2n−2p−1)

"

2²ⁿ⁻² −n

p

X

k=0

2n k

#

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= 2²ⁿ⁻²n²−n (_n−1

X

p=0

(2n−2p−1)

p

X

k=0

2n k

)

= 2²ⁿ⁻²n²−n

n−1

X

k=0

(n−k)² 2n

k

= 0.

This implies that inequality (3.3) is equivalent to

(3.4)

2n

X

i=1,j=1,i6=j

(x_i−x_j)² ( _n

X

k=1

"

k2²ⁿ⁻²−n

k−1

X

q=0

(k−q) 2n

q #

x^2n−k−1_i x^k−1_j

+

2n

X

k=n+1

"

(2n−k+ 1)2²ⁿ⁻²−n

2n

X

q=k

(2n−q+ 1) 2n

q−k #

x^2n−k_i x^k−2_j )

≥0,

2n

X

i=1,j=1,i6=j

(_n−1 X

k=1

"

k(k+ 1)

2 2²ⁿ⁻²−n

k

X

p=1

p(p+ 1) 2

2n k−p

#

×x^k−1_i x^k−1_j

2n−2k−2

X

p=0

x^2n−p−4_i x^p_j )

(x_i−x_j)⁴ ≥0.

In order to prove (3.4), it is sufficient to show (3.5) (n−1)[(n−1) + 1]

2 2²ⁿ⁻²−n

n−1

X

p=1

p(p+ 1) 2

2n n−p−1

>0.

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Considering (2.1), it is sufficient to show (3.6)

n−1

X

k=0

(n−k) 2n

k

>2²ⁿ⁻².

By virtue of ⁿ_k

= _n−kⁿ

andP2n k=0

2n k

= 2²ⁿ, inequality (3.6) can be rearranged as

n−1

X

k=0

(n−k) 2n

k

+

2n

X

k=n+1

(k−n) 2n

k

>2²ⁿ⁻¹,

n−1

X

k=0

(2n−2k−1) 2n

k

+

2n

X

k=n+1

(2k−2n−1) 2n

k

>

2n n

. (3.7)

Sincen ≥ 2and _n−1²ⁿ

+ _n+1²ⁿ

> ²ⁿ_n

is equivalent to 2 > ⁿ⁺¹_n , then inequalities (3.7), (3.6) and (3.5) are valid. The proof of Theorem1.1is complete.

Proof of Theorem1.2. By Lemma 2.2, it is easy to see that P2n

i=1y_i = Pn i=1x_i. From Cauchy’s inequality [1,4], it follows that

A_n

2n

X

i=1

y²_i yi−1|2nPi+n−2

k=i yk|2n

≥

2n

X

i=1

y_i

!2

,

where A_n is defined by (2.2) or (2.3) in Lemma 2.2. Therefore, it is sufficient to

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prove (n−1)

2n

X

i=1

yi

!2

≥2nAn⇐⇒ (n−1)

n

X

i=1

xi

!2

≥2n X

1≤i<j≤n

xixj

⇐⇒ (n−1)

n

X

i=1

x²_i ≥2 X

1≤i<j≤2n

x_ix_j

⇐⇒ X

1≤i<j≤2n

(x_i−x_j)² ≥0.

The proof of Theorem1.2is complete.

Proof of Theorem1.3. Let

(3.8) B_n = 2²⁽ⁿ⁻¹⁾k(k+ 1)−n

k

X

p=1

p(p+ 1) 2n

k−p

.

Then

B_n+1 =k(k+ 1)2²ⁿ⁻²2²−(n+ 1)

k

X

p=1

p(p+ 1)

2n+ 2 k−p

= 4B_n+

k

X

p=1

p(p+ 1)

4n 2n

k−p

−(n+ 1)

2n+ 2 k−p

,4Bn+

k

X

p=1

p(p+ 1)Ck−p

(3.9)

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and

C_q−C_q+1 = 4n 2n

q

− 2n

q+ 1

−(n+ 1)

2n+ 2 q

−

2n+ 2 q+ 1

= 4n 2n

q 1− 2n−q q+ 1

−(n+ 1)

2n+ 2

q 1− 2n+ 2−q q+ 1

= 4n 2n

q

2q−2n+ 1

q+ 1 −(n+ 1)

2n+ 2 q

2q−2n−1 q+ 1

> 2q−2n+ 1 q+ 1 C_q for0≤q≤k−1. Hence,

(3.10) 2n−q

q+ 1 C_q> C_q+1. From the above inequality and the facts that

(3.11) C_n = 2(2n−1)(n+ 1)

n+ 2

2n n

>0

and ^2n−q_q+1 > 0, it follows easily thatC_q > 0. Consequently, we haveB_n+1 > 4B_n, and thenB_k+2 >4B_k+1. As a result, utilization of (3.5) gives

Bk+1 >0, Bk+2 >0, Bk+3 >0, Bk+4 >0, · · · , Bk+(n−k) =Bn>0.

The proof of inequality (1.3) is complete.

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References

[1] J.-CH. KUANG, Chángyòng Bùdˇengshì (Applied Inequalities), 3rd ed., Sh¯and¯ong K¯exué Jìshù Ch¯ubˇan 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. MITRINOVI ´C, J. E. PE ˇCARI ´C, 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)