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volume 7, issue 3, article 107, 2006.

Received 15 March, 2006;

accepted 04 April, 2006.

Communicated by:J. Pecari´c

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Journal of Inequalities in Pure and Applied Mathematics

EXTENSIONS OF SEVERAL INTEGRAL INEQUALITIES

FENG QI, AI-JUN LI, WEI-ZHEN ZHAO, DA-WEI NIU AND JIAN CAO

Research Institute of Mathematical Inequality Theory Henan Polytechnic University

Jiaozuo City, Henan Province, 454010, China.

EMail:qifeng@hpu.edu.cn URL:http://rgmia.vu.edu.au/qi.html

School of Mathematics and Informatics Henan Polytechnic University

Jiaozuo City, Henan Province 454010, China

EMail:liaijun72@163.com EMail:zhao_weizhen@sina.com EMail:nnddww@tom.com EMail:21caojian@163.com

c

2000Victoria University ISSN (electronic): 1443-5756 073-06

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Extensions of Several Integral Inequalities

Feng Qi, Ai-Jun Li, Wei-Zhen Zhao, Da-Wei Niu and

Jian Cao

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J. Ineq. Pure and Appl. Math. 7(3) Art. 107, 2006

Abstract

In this article, an open problem posed in [12] is studied once again, and, following closely theorems and methods from [5], some extensions of several integral inequalities are obtained.

2000 Mathematics Subject Classification:Primary 26D15.

Key words: Integral inequality, Cauchy’s Mean Value Theorem.

The authors were supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China.

The authors would like to heartily express their thanks to the anonymous referee for his/her valuable comments and corrections to this paper.

1. Introduction

In [12], the following interesting integral inequality is proved: Let f(x) be continuous on [a, b] and differentiable on (a, b) such that f(a) = 0. If 0 ≤ f⁰(x)≤1forx∈(a, b), then

(1.1)

Z b

a

[f(x)]³dx≤ Z b

a

f(x) dx ²

.

Iff⁰(x) ≥ 1, then inequality (1.1) reverses. The equality in (1.1) holds only if f(x)≡0orf(x) =x−a.

As a generalization of inequality (1.1), the following more general result is also obtained in [12]: Letn ∈Nand supposef(x)has a continuous derivative of then-th order on the interval[a, b]such thatf⁽ⁱ⁾(a) ≥ 0for0≤ i ≤ n−1 andf⁽ⁿ⁾(x)≥n!. Then

(1.2)

Z b

a

[f(x)]ⁿ⁺²dx≥ Z b

a

f(x) dx ⁿ⁺¹

.

At the end of [12] an open problem is proposed: Under what conditions does the inequality

(1.3)

Z b

a

f(x)t

dx≥ Z b

a

f(x) dx ^t−1

hold fort >1?

This open problem has attracted some mathematicians’ research interests and many generalizations, extensions and applications of inequality (1.2) or

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(1.3) were investigated in recent years. For more detailed information, please refer to, for example, [1,2, 3,4, 5,6, 7,8, 9,10, 11,13, 14,15] and the references therein.

In this paper, following closely theorems and methods from [5], we will establish some more extensions and generalizations of inequality (1.2) or (1.3) once again. Our main results are the following five theorems.

Theorem 1.1. Letf(x)be continuous and not identically zero on[a, b], differ- entiable in(a, b), withf(a) = 0, and letα, βbe positive real numbers such that α > β > 1. If

(1.4)

f(α−β)/(β−1)

(x)0

R (α−β)β^1/(β−1) α−1 for allx∈(a, b), then

(1.5)

Z b

a

[f(t)]^αdtR Z b

a

f(t) dt ^β

.

Theorem 1.2. Letα∈Randf(x)be continuous on[a, b]and positive in(a, b).

1. Forβ >1, if (1.6)

Z x

a

f(t) dtQβ^1/(1−β)[f(x)](α−1)/(β−1)

for allx∈(a, b), then inequality (1.5) is validated;

2. For0< β <1, if inequality (1.6) is reversed, then inequality (1.5) holds;

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3. For β = 1, if [f(x)]^1−α Q 1 for all x ∈ (a, b), then inequality (1.5) is valid.

Theorem 1.3. Suppose n ∈ N, 1 ≤ β ≤ n+ 1, andf(x)has a derivative of then-th order on the interval[a, b]such thatf⁽ⁱ⁾(a) = 0for0≤i≤n−1and f⁽ⁿ⁾(x)≥0.

1. Iff(x)≥ h_(x−a)β−1

β^β−2

i^1/(α−β)

andf⁽ⁿ⁾(x)is increasing, then the inequality with direction≥in (1.5) holds.

2. If 0 ≤ f(x) ≤ h_(x−a)β−1

β^β−2

i1/(α−β)

andf⁽ⁿ⁾(x) is decreasing, then the in- equality with direction≤in (1.5) is valid.

Theorem 1.4. Suppose n ∈ N, 1 < β ≤ n+ 1, andf(x)has a derivative of then-th order on the interval[a, b]such thatf⁽ⁱ⁾(a) = 0for0≤i≤n−1and f⁽ⁿ⁾(x)≥0.

1. Iff(x)≥h_β(x−a)(β−1)

(β−1)^(β−1)

i1/(α−β)

, then the inequality with direction≥in (1.5) holds.

2. If0≤f(x)≤h_β(x−a)(β−1)

(β−1)^(β−1)

i1/(α−β)

, then the inequality with direction≤in (1.5) is valid.

Theorem 1.5. Letα, βbe positive numbers,α > β ≥2andf(x)be continuous on[a, b]and differentiable on(a, b)such thatf(a)≥0. If

[f^(α−β)(x)]⁰ ≥ β(β−1)(α−β)(x−a)^β−2 α−1

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forx∈(a, b),then the inequality with direction≥in (1.5) holds.

Remark 1. Theorem 1.5 generalizes a result obtained in [9, Theorem 2] by Peˇcari´c and Pejkovi´c.

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2. Proofs of Theorems

Proof of Theorem1.1. If

f(α−β)/(β−1)

(x)⁰

≥ (α−β)β^1/(β−1) α−1

forx∈ (a, b)andα > β >1, thenf(x) >0forx∈ (a, b]. Thus both sides of (1.5) do not equal zero. This allows us to consider the quotient of both sides of (1.5). Utilizing Cauchy’s Mean Value Theorem consecutively yields

hRb

af(t) dtiβ

Rb

a[f(t)]^αdt = βh

Rξ

a f(t) dtiβ−1

f(ξ)

[f(ξ)]^α ξ ∈(a, b)

=

(β^1/(β−1)Rξ

a f(t) dt [f(ξ)](α−1)/(β−1)

)β−1

(2.1)

=

( β^1/(β−1)f(θ)

α−1

β−1[f(θ)](α−β)/(β−1)f⁰(θ) )^β−1

θ ∈(a, ξ)

=

((α−β)β^1/(β−1)/(α−1) f(α−β)/(β−1)(θ)0

)β−1

(2.2)

≤1.

So the inequality with direction≥in (1.5) follows.

If

0≤

f(α−β)/(β−1)

(x)0

≤ (α−β)β^1/(β−1) α−1

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for x ∈ (a, b) and α > β > 1, then f(α−β)/(β−1)(x) is nondecreasing and f(x) ≥ 0forx ∈ [a, b]. Without loss of generality, we may assumef(x) > 0 for x ∈ (a, b] (otherwise, we can find a pointa₁ ∈ (a, b)such that f(a₁) = 0 andf(x)>0forx∈(a₁, b]and hence we only need to consider the inequality with direction ≤ in (1.5) on [a₁, b]). This means that both sides of inequality (1.5) are not zero. Therefore, the inequality with direction ≤in (1.5) follows from (2.2).

Proof of Theorem1.2. The first and second conclusions are obtained easily by (2.1) of Theorem1.1.

Forβ = 1, inequality (1.5) is reduced to (2.3)

Z b

a

[f(t)]^αdtR Z b

a

f(t) dt.

Now consider the quotient of both sides of (2.3). By Cauchy’s Mean Value Theorem, it is obtained that

(2.4)

Rb

a[f(t)]^αdt Rb

af(t) dt = [f(ξ)]^α

f(ξ) = [f(ξ)]^α−1.

The third conclusion is proved.

Proof of Theorem1.3. Utilization of the condition thatf(x)≥h_(x−a)β−1

β^β−2

i^1/(α−β)

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and Cauchy’s Mean Value Theorem gives Rb

a[f(x)]^αdx Rb

a f(x) dxβ = [f(b₁)]^α−1 β Rb1

a f(x) dxβ−1 a < b₁ < b (2.5)

≥ (b₁−a)^β−1[f(b₁)]^β−1 β^β−2 β Rb1

a f(x) dxβ−1

(2.6)

=

"

(b₁−a)f(b₁) βRb1

a f(x) dx

#β−1

. (2.7)

Now for the term in (2.7), by using Cauchy’s Mean Value Theorem several times, we have

(b₁−a)f(b₁) Rb1

a f(x) dx = 1 + (b₂−a)f⁰(b₂)

f(b₂) a < b₂ < b₁

= 2 + (b3−a)f⁰⁰(b3)

f⁰(b₃) a < b₃ < b₂

· · ·

=n+(bn+1−a)f⁽ⁿ⁾(bn+1)

f⁽ⁿ⁻¹⁾(b_n+1) a < b_n+1 < b_n. (2.8)

Butf⁽ⁿ⁻¹⁾(t) =f⁽ⁿ⁻¹⁾(t)−f⁽ⁿ⁻¹⁾(a) = (t−a)f⁽ⁿ⁾(t₁)for somet₁ ∈(a, t). If f⁽ⁿ⁾(x)is increasing, thenf⁽ⁿ⁾(t₁)≤f⁽ⁿ⁾(t). Therefore,

(2.9) 0< f⁽ⁿ⁻¹⁾(t)≤f⁽ⁿ⁾(t)(t−a).

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Applying (2.9) to (2.8) yields

(2.10) (b₁−a)f(b₁)

Rb1

a f(x) dx ≥n+ 1.

Hence, (2.11)

Rb

a[f(x)]^αdx Rb

af(x) dxβ ≥

n+ 1 β

β−1

for1≤β ≤n+ 1. Then the inequality with direction≥in (1.5) holds.

Suppose that

0≤f(x)≤h(x−a)^β−1 β^β−2

i1/(α−β)

and f⁽ⁿ⁾(x) is decreasing. The statement of the theorem implies that the inequalities (2.6) and (2.9) reverse, this means that the inequalities (2.10) and (2.11) reverse also, so the inequality with direction≤in (1.5) holds.

Proof of Theorem1.4. If

f(x)≥hβ(x−a)^(β−1) (β−1)^(β−1)

i1/(α−β)

,

(2.5) becomes Rb

a[f(x)]^αdx hRb

a f(x) dxiβ ≥

"

(b₁−a)f(b₁) (β−1)Rb1

a f(x) dx

#β−1

.

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Note that if all the terms in (2.8) are positive, then ^(bR¹_b₁^−a)f(b¹⁾

a f(x) dx ≥ n. Therefore, for1< β≤n+ 1, the inequality with direction≥in (1.5) holds.

If

0≤f(x)≤hβ(x−a)^(β−1) (β−1)^(β−1)

i1/(α−β)

,

the inequality with direction ≤ in (1.5) follows from a similar argument as above.

Proof of Theorem1.5. Suppose that

[f^(α−β)(x)]⁰ ≥ β(β−1)(α−β)(x−a)^β−2

α−1 .

Now consider the quotient of the two sides of (1.5). Applying Cauchy’s Mean Value Theorem three times leads to

Rb

a[f(x)]^αdx Rb

af(x) dxβ = [f(b₁)]^α−1 β Rb1

a f(x) dxβ−1

≥ (α−1)[f(b2)]^α−3f⁰(b2) β(β−1) Rb2

a f(x) dxβ−2 a < b₂ < b₁

≥

"

f(b2)(b2−a) Rb2

a f(x) dx

#β−2

a < b₃ < b₂

=

1 + f⁰(b3)(b3−a) f(b₃)

β−2

≥1.

This completes the proof.

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