In this article, a discrete version of an open problem in [Q

A DISCRETE VERSION OF AN OPEN PROBLEM AND SEVERAL ANSWERS

YU MIAO AND FENG QI

COLLEGE OFMATHEMATICS ANDINFORMATIONSCIENCE

HENANNORMALUNIVERSITY

XINXIANGCITY, HENANPROVINCE

453007, CHINA

yumiao728@yahoo.com.cn

RESEARCHINSTITUTE OFMATHEMATICALINEQUALITYTHEORY

HENANPOLYTECHNICUNIVERSITY

JIAOZUOCITY, HENANPROVINCE

454010, CHINA

qifeng618@gmail.com

Received 10 July, 2008; accepted 29 May, 2009 Communicated by S.S. Dragomir

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 athttp://jipam.vu.edu.au/article.php?

sid=737] is posed and several answers are provided.

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

2000 Mathematics Subject Classification. 26D15; 28A25.

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?

199-08

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

In [3], an abstract version of Open Problem 1 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. For n ∈ N, let {x₁, x₂, . . . , x_n} and {y₁, y₂, . . . , y_n} be two positive se- quences satisfyingx₁ ≥x₂ ≥ · · · ≥x_n,y₁ ≥y₂ ≥ · · · ≥y_nand

(1.3)

m

X

i=1

x_i ≤

m

X

i=1

y_i

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

(1.4)

n

X

i=1

x^α_iy_i^β ≤

n

X

i=1

y_i^α+β

hold forαandβ?

In the next sections, we shall establish several answers to Open Problem 2.

2. LEMMAS

In order to establish several answers to Open Problem 2, 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. For n ∈ N, let {x₁, x₂, . . . , x_n} and {y₁, y₂, . . . , y_n}be two positive sequences satisfyingx1 ≥x2 ≥ · · · ≥xn,y1 ≥y2 ≥ · · · ≥ynand 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. Let x_n+1 be a positive number such that x_n+1 ≤ x_n. From Lemma 2.1 and using in- equality (1.3), it is easy to see that, forα= 2and1≤m≤n,

m

X

i=1

x_iy_i =x_m+1

m

X

i=1

y_i+

m

X

i=1 i

X

j=1

y_j(x_i−x_i+1)

≥x_m+1

m

X

i=1

x_i +

m

X

i=1 i

X

j=1

x_j(x_i−x_i+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 integer α > 2. Since {x₁, x₂, . . . , x_n} and {y₁, y₂, . . . , y_n}are two positive sequences, then

(y^α_i −x^α_i)(yi−xi)≥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 Lemma 2.1, it follows that

m

X

i=1

y_i^αx_i =x_m+1

m

X

i=1

y_i^α+

m

X

i=1 i

X

j=1

y_j^α(x_i−x_i+1)

≥x_m+1

m

X

i=1

x^α_i +

m

X

i=1 i

X

j=1

x^α_j(x_i−x_i+1)

=

m

X

i=1

x^α+1_i . (2.6)

A similar argument also yields (2.7)

m

X

i=1

yix^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 numberα ≥ 1. By inequality (2.2) in Lemma 2.2, we have

(2.8) [α]

α y^α_i + α−[α]

α x^α_i ≥y_i^[α]x^α−[α]_i .

Summing on both sides of (2.8) and utilizing Lemma 2.1, the conclusion obtained above for α∈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.

3. SEVERAL ANSWERS TOOPENPROBLEM2 Now we establish several answers to Open Problem 2.

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

(3.1)

n

X

i=1

x^α_iy_i^β ≤

n

X

i=1

y_i^α+β

holds forα≥1andβ >0.

Proof. By Hölder’s inequality and Lemma 2.3,

n

X

i=1

x^α_iy_i^β ≤

" _n X

i=1

(x^α_i)^α+βα

#_α+β^α " _n X

i=1

y_i^βα+β_β

#_α+β^β

≤ Pn

i=1x^α+β_i Pn

i=1y^α+β_i

!_α+β^α _n X

i=1

y_i^α+β

≤

n

X

i=1

y^α+β_i .

This completes the proof of Theorem 3.1.

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 >0and1≤l ≤k be positive sequences such thatx_1,l ≥x_2,l ≥ · · · ≥x_n,l,y_1,l ≥y_2,l ≥ · · · ≥y_n,l and

(3.2)

m

X

i=1

xi,l≤

m

X

i=1

yi,l, 1≤m≤n, 1≤l ≤k.

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^αl+βl forα_l≥1andβ_l>0,1≤l ≤k.

Proof. As in the proof of Lemma 2.3, letx_n+1,l be positive numbers such thatx_n+1,l ≤x_n,lfor 1≤l≤k. By Lemma 2.1 and Theorem 3.1, 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^βl

n

X

i=1

x^α_i,k^ky_i,k^βk

+

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^βl−

k−1

Y

l=1

x^α_i+1,l^l y_i+1,l^βl

!

≤

k−1

Y

l=1

x^α_n+1,l^l y_n+1,l^βl

n

X

i=1

y^α_i,k^k+βk

+

n

X

i=1 i

X

j=1

y_j,k^αk+βk

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^βl

!

=

n

X

i=1

y^α_j,k^k+βk

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+βl.

The proof of Theorem 3.2 is completed.

Theorem 3.3. Forn ∈ N, let{x₁, x₂, . . . , x_n}and{y₁, y₂, . . . , y_n} be two positive sequences with the properties that x₁ ≥ x₂ ≥ · · · ≥ x_n, y₁ ≥ y₂ ≥ · · · ≥ y_n and inequality (1.3) is satisfied. Then

(3.4)

n

X

i=1

y_i^α1x^β_i¹ ≤

n

X

i=1

y_i^αx^β_i

ifα ≥α₁ ≥1,β >0andβ+α =β₁+α₁.

Proof. Letx_n+1 be a positive number such thatx_n+1 ≤ x_n. By Lemma 2.1 and Theorem 3.1, 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_j^α 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 Theorem 3.3.

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. LITTLEWOODANDG. PÓLYA, Inequalities, 2nd edition, Cambridge Univer- sity Press, Cambridge, 1952.

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

[4] Q.A. NGÔ, D.D. THANG, T.T. DATANDD.A. TUAN, Notes on an integral inequality, J. Inequal.

Pure Appl. Math., 7(4) (2006), Art. 120. [ONLINE:http://jipam.vu.edu.au/article.

php?sid=737].