A DISCRETE VERSION OF AN OPEN PROBLEM AND SEVERAL ANSWERS

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Open Problem Yu Miao and Feng Qi vol. 10, iss. 2, art. 49, 2009

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A DISCRETE VERSION OF AN OPEN PROBLEM AND SEVERAL ANSWERS

YU MIAO FENG QI

College of Mathematics and Information Science Research Instit. of Mathematical Inequality Theory Henan Normal University Henan Polytechnic University

Xinxiang City, Henan Province Jiaozuo City, Henan Province

453007, China 454010, China

EMail:yumiao728@yahoo.com.cn EMail:qifeng618@gmail.com

Received: 10 July, 2008

Accepted: 29 May, 2009

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15; 28A25

Key words: Integral inequality, Discrete version, Open problem.

Abstract: In this article, a discrete version of an open problem in [Q. A. Ngô, D. D.

Thang, T. T. Dat, and D. A. Tuan, Notes on an integral inequality, J. Inequal.

Pure Appl. Math. 7 (2006), no. 4, Art. 120; Available online at http:

//jipam.vu.edu.au/article.php?sid=737] is posed and several answers are provided.

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1. Introduction

In [4], some integral inequalities were obtained and the following open problem was posed.

Open Problem 1. Letf be a continuous function on[0,1]satisfying the following condition

(1.1)

Z 1

x

f(t) dt≥ Z 1

x

tdt

forx∈[0,1]. Under what conditions does the inequality

(1.2)

Z 1

0

f^α+β(t) dt ≥ Z 1

0

t^βf^α(t) dt

hold forαandβ?

In [1], some affirmative answers to Open Problem1and the reversed inequality of (1.2) were given.

In [3], an abstract version of Open Problem1 was posed, respective answers to these two open problems were presented, and the results in [1] were extended.

Now we would like to further pose the following discrete version of the open problems in [1,3] as follows.

Open Problem 2. Forn∈N, let{x₁, x₂, . . . , x_n}and{y₁, y₂, . . . , y_n}be two posi- tive sequences satisfyingx₁ ≥x₂ ≥ · · · ≥x_n,y₁ ≥y₂ ≥ · · · ≥y_nand

(1.3)

m

X

i=1

x_i ≤

m

X

i=1

y_i

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for1≤m≤n. Under what conditions does the inequality

(1.4)

n

X

i=1

x^α_iy^β_i ≤

n

X

i=1

y_i^α+β hold forαandβ?

In the next sections, we shall establish several answers to Open Problem2.

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2. Lemmas

In order to establish several answers to Open Problem2, the following lemmas are necessary.

Lemma 2.1. Forn ∈ N, let{x₁, x₂, . . . , x_n, x_n+1}and{y₁, y₂, . . . , y_n}be two real sequences. Then

(2.1)

n

X

i=1

x_iy_i =x_n+1

n

X

i=1

y_i+

n

X

i=1 i

X

j=1

y_j(x_i−x_i+1).

Proof. Identity (2.1) follows from standard straightforward arguments.

Lemma 2.2 ([2, p. 17]). Letaandbbe positive numbers witha+b= 1. Then

(2.2) ax+by ≥x^ay^b

is valid for positive numbersxandy.

Lemma 2.3. Forn ∈ N, let{x₁, x₂, . . . , x_n}and {y₁, y₂, . . . , y_n}be two positive sequences satisfyingx₁ ≥x₂ ≥ · · · ≥ x_n, y₁ ≥y₂ ≥ · · · ≥y_nand inequality (1.3).

Then

(2.3)

m

X

i=1

x^α_i ≤

m

X

i=1

y_i^α holds forα≥1and1≤m≤n.

Proof. Letxn+1 be a positive number such thatxn+1 ≤ xn. From Lemma2.1 and

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using inequality (1.3), it is easy to see that, forα= 2and1≤m ≤n,

m

X

i=1

xiyi =xm+1 m

X

i=1

yi+

m

X

i=1 i

X

j=1

yj(xi−xi+1)

≥xm+1 m

X

i=1

xi+

m

X

i=1 i

X

j=1

xj(xi−xi+1)

=

m

X

i=1

x²_i

which implies that (2.4)

m

X

i=1

y_i² ≥2

m

X

i=1

x_iy_i−

m

X

i=1

x²_i ≥

m

X

i=1

x²_i.

Suppose that inequality (2.3) holds for some integerα >2. Since{x₁, x₂, . . . , x_n} and{y₁, y₂, . . . , y_n}are two positive sequences, then

(y_i^α−x^α_i)(y_i−x_i)≥0 which leads to

(2.5)

m

X

i=1

y^α+1_i ≥

m

X

i=1

y^α_ix_i+

m

X

i=1

y_ix^α_i −

m

X

i=1

x^α+1_i

for1≤m≤n. Further, by virtue of Lemma2.1, it follows that

m

X

i=1

y_i^αx_i =x_m+1

m

X

i=1

y^α_i +

m

X

i=1 i

X

j=1

y_j^α(x_i−x_i+1) (2.6)

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≥xm+1 m

X

i=1

x^α_i +

m

X

i=1 i

X

j=1

x^α_j(xi −xi+1) =

m

X

i=1

x^α+1_i .

A similar argument also yields (2.7)

m

X

i=1

y_ix^α_i ≥

m

X

i=1

x^α+1_i .

Substituting (2.6) and (2.7) into (2.5) gives inequality (2.3) forα+ 1.

By induction, this means that inequality (2.3) holds for allα∈N.

Let [α] denote the integral part of a real number α ≥ 1. By inequality (2.2) in Lemma2.2, we have

(2.8) [α]

α y_i^α+ α−[α]

α x^α_i ≥y_i^[α]x^α−[α]_i .

Summing on both sides of (2.8) and utilizing Lemma2.1, the conclusion obtained above forα∈Nyields

[α]

α

m

X

i=1

y_i^α ≥

m

X

i=1

y_i^[α]x^α−[α]_i − α−[α]

α

m

X

i=1

x^α_i

=x^α−[α]_m+1

m

X

i=1

y_i^[α]+

m

X

i=1 i

X

j=1

y^[α]_j

x^α−[α]_i −x^α−[α]_i+1

−α−[α]

α

m

X

i=1

x^α_i

≥

m

X

i=1

x^α_i −α−[α]

α

m

X

i=1

x^α_i = [α]

α

m

X

i=1

x^α_i.

Since ^[α]_α 6= 0, the required result is proved.

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3. Several Answers to Open Problem 2

Now we establish several answers to Open Problem2.

Theorem 3.1. Forn ∈ N, let{x₁, x₂, . . . , x_n}and{y₁, y₂, . . . , y_n}be two positive sequences such thatx₁ ≥x₂ ≥ · · · ≥ x_n,y₁ ≥y₂ ≥ · · · ≥ y_nand inequality (1.3) is satisfied. Then

(3.1)

n

X

i=1

x^α_iy^β_i ≤

n

X

i=1

y_i^α+β

holds forα≥1andβ >0.

Proof. By Hölder’s inequality and Lemma2.3,

n

X

i=1

x^α_iy_i^β ≤

" _n X

i=1

(x^α_i)^α+β^α

#_α+β^α " _n X

i=1

y_i^β^α+β_β

#_α+β^β

≤ Pn

i=1x^α+β_i Pn

i=1y_i^α+β

!_α+β^α _n X

i=1

y_i^α+β ≤

n

X

i=1

y^α+β_i .

This completes the proof of Theorem3.1.

Theorem 3.2. Let{x_1,l, x_2,l, . . . , x_n,l}and{y_1,l, y_2,l, . . . , y_n,l}forn∈N,k >0and 1 ≤ l ≤ k be positive sequences such that x_1,l ≥ x_2,l ≥ · · · ≥ x_n,l, y_1,l ≥ y_2,l ≥

· · · ≥y_n,l and

(3.2)

m

X

i=1

x_i,l ≤

m

X

i=1

y_i,l, 1≤m≤n, 1≤l ≤k.

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Then

(3.3)

n

X

i=1 k

Y

l=1

x^α_i,l^ly^β_i,l^l ≤

n

X

i=1 k

Y

l=1

y_i,l^α^l^+β^l

forα_l ≥1andβ_l>0,1≤l ≤k.

Proof. As in the proof of Lemma2.3, letx_n+1,lbe positive numbers such thatx_n+1,l ≤ x_n,lfor1≤l ≤k. By Lemma2.1and Theorem3.1, it is shown that

n

X

i=1 k

Y

l=1

x^α_i,l^ly^β_i,l^l =

k−1

Y

l=1

x^α_n+1,l^l y_n+1,l^β^l

n

X

i=1

x^α_i,k^ky_i,k^β^k

+

n

X

i=1 i

X

j=1

x^α_j,k^ky^β_j,k^k

k−1

Y

l=1

x^α_i,l^ly_i,l^β^l−

k−1

Y

l=1

x^α_i+1,l^l y_i+1,l^β^l

!

≤

k−1

Y

l=1

x^α_n+1,l^l y_n+1,l^β^l

n

X

i=1

y_i,k^α^k^+β^k

+

n

X

i=1 i

X

j=1

y_j,k^α^k^+β^k

k−1

Y

l=1

x^α_i,l^ly^β_i,l^l−

k−1

Y

l=1

x^α_i+1,l^l y_i+1,l^β^l

!

=

n

X

i=1

y^α_j,k^k^+β^k

k−1

Y

l=1

x^α_i,l^ly^β_i,l^l ≤ · · · ≤

n

X

i=1 k

Y

l=1

y^α_i,l^l^+β^l.

The proof of Theorem3.2is completed.

Theorem 3.3. Forn ∈ N, let{x₁, x₂, . . . , x_n}and{y₁, y₂, . . . , y_n}be two positive sequences with the properties thatx₁ ≥ x₂ ≥ · · · ≥ x_n, y₁ ≥ y₂ ≥ · · · ≥ y_n and

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inequality (1.3) is satisfied. Then

(3.4)

n

X

i=1

y_i^α¹x^β_i¹ ≤

n

X

i=1

y^α_ix^β_i

ifα ≥α₁ ≥1,β >0andβ+α=β₁+α₁.

Proof. Let x_n+1 be a positive number such that x_n+1 ≤ x_n. By Lemma 2.1 and Theorem3.1, we have

n

X

i=1

y_i^αx^β_i =x^β_n+1

n

X

i=1

y_i^α+

n

X

i=1 i

X

j=1

y^α_j x^β_i −x^β_i+1

≥ x^β_n+1

n

X

i=1

y^α_i¹x^α−α_i ¹ +

n

X

i=1 i

X

j=1

y^α_j¹x^α−α_j ¹ x^β_i −x^β_i+1

=

n

X

i=1

y_i^α¹x^α−α_i ¹^+β =

n

X

i=1

y_i^α¹x^β_i¹ which completes the proof of Theorem3.3.

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References

[1] L. BOUGOFFA, Note on an open problem, J. Inequal. Pure Appl. Math., 8(2) (2007), Art. 58. [ONLINE:http://jipam.vu.edu.au/article.php?

sid=871].

[2] G.H. HARDY, J.E. LITTLEWOOD ANDG. PÓLYA, Inequalities, 2nd edition, Cambridge University Press, Cambridge, 1952.

[3] Y. MIAOANDF. QI, Another answer to an open problem, submitted.

[4] Q.A. NGÔ, D.D. THANG, T.T. DAT AND D.A. TUAN, Notes on an inte- gral inequality, J. Inequal. Pure Appl. Math., 7(4) (2006), Art. 120. [ONLINE:

http://jipam.vu.edu.au/article.php?sid=737].