ON AN OPEN PROBLEM CONCERNING AN INTEGRAL INEQUALITY

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Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li

and Jian-wei Dong vol. 8, iss. 3, art. 74, 2007

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ON AN OPEN PROBLEM CONCERNING AN INTEGRAL INEQUALITY

WEN-JUN LIU, CHUN-CHENG LI

College of Mathematics and Physics

Nanjing University of Information Science and Technology Nanjing, 210044 China

EMail:[email protected] [email protected]

JIAN-WEI DONG

Department of Mathematics and Physics

Zhengzhou Institute of Aeronautical Industry Management Zhengzhou 450015, China

EMail:[email protected]

Received: 09 December, 2006

Accepted: 11 August, 2007

Communicated by: P.S. Bullen 2000 AMS Sub. Class.: 26D15.

Key words: Integral inequality, Cauchy inequality.

Abstract: In this note, we generalize an open problem posed by Q. A. Ngô in the paper, Notes on an integral inequality, J. Inequal. Pure & Appl. Math., 7(4)(2006), Art. 120 and give a positive answer to it using an analytic approach.

Acknowledgements: The first author was supported by the Science Research Foundation of NUIST and the Natural Science Foundation of Jiangsu Province Education Department under Grant No.07KJD510133, and the third author was supported by Youth Natural Science Foun- dation of Zhengzhou Institute of Aeronautical Industry Management under Grant No.

Q05K066.

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1. Introduction

In the paper [2], Q.A. Ngô studied a very interesting integral inequality and proved the following result.

Theorem 1.1. Letf(x)≥0be a continuous function on[0,1]satisfying (1.1)

Z 1

x

f(t)dt ≥ Z 1

x

t dt, ∀x∈[0,1].

Then the inequalities (1.2)

Z 1

0

f^α+1(x)dx≥ Z 1

0

x^αf(x)dx and

(1.3)

Z 1

0

f^α+1(x)dx≥ Z 1

0

xf^α(x)dx

hold for every positive real numberα >0.

Next, they proposed the following open problem.

Problem 1. Letf(x)be a continuous function on[0,1]satisfying (1.4)

Z 1

x

f(t)dt ≥ Z 1

x

t dt, ∀x∈[0,1].

Under what conditions does the inequality (1.5)

Z 1

0

f^α+β(x)dx≥ Z 1

0

x^αf^β(x)dx, hold forαandβ?

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We note that, as an open problem, the condition (1.4) maybe result in an unrea- sonable restriction onf(x). We remove it herein and propose another more general open problem.

Problem 2. Under what conditions does the inequality (1.6)

Z b

0

f^α+β(x)dx≥ Z b

0

x^αf^β(x)dx, hold forb, αandβ?

In this note, we give an answer to Problem 2 using an analytic approach. Our main results are Theorem2.1and Theorem2.4which will be proved in Section2.

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2. Main Results and Proofs

Firstly, we have

Theorem 2.1. Letf(x)≥0be a continuous function on[0,1]satisfying (2.1)

Z 1

x

f^β(t)dt≥ Z 1

x

t^βdt, ∀x∈[0,1].

Then the inequality (2.2)

Z 1

0

f^α+β(x)dx≥ Z 1

0

x^αf^β(x)dx,

holds for every positive real numberα >0andβ >0.

To prove Theorem2.1, we need the following lemmas.

Lemma 2.2 (General Cauchy inequality, [2]). Letαandβbe positive real numbers satisfyingα+β = 1. Then for all positive real numbersxandy, we have

(2.3) αx+βy ≥x^αy^β.

Lemma 2.3. Under the conditions of Theorem2.1, we have (2.4)

Z 1

0

x^αf^β(x)dx≥ 1 α+β+ 1.

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Proof. Integrating by parts, we have Z 1

0

x^α−1 Z 1

x

f^β(t)dt

dx= 1 α

Z 1

0

Z 1

x

f^β(t)dt

d(x^α) (2.5)

= 1 α

x^α

Z 1

x

f^β(t)dt ^x=1

x=0

+ 1 α

Z 1

0

x^αf^β(x)dx

= 1 α

Z 1

0

x^αf^β(x)dx,

which yields (2.6)

Z 1

0

x^αf^β(x)dx=α Z 1

0

x^α−1 Z 1

x

f^β(t)dt

dx.

On the other hand, by (2.1), we get Z 1

0

x^α−1 Z 1

x

f^β(t)dt

dx≥ Z 1

0

x^α−1 Z 1

x

t^βdt

dx (2.7)

= 1

β+ 1 Z 1

0

(x^α−1−x^α+β)dx

= 1

α(α+β+ 1). Therefore, (2.4) holds.

We now give the proof of Theorem2.1.

Proof of Theorem2.1. Using Lemma2.2, we obtain

(2.8) β

α+βf^α+β(x) + α

α+βx^α+β ≥x^αf^β(x),

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which gives

(2.9) β

Z 1

0

f^α+β(x)dx+α Z 1

0

x^α+βdx≥(α+β) Z 1

0

x^αf^β(x)dx.

Moreover, by using Lemma2.3, we get (α+β)

Z 1

0

x^αf^β(x)dx=α Z 1

0

x^αf^β(x)dx+β Z 1

0

x^αf^β(x)dx (2.10)

≥ α

α+β+ 1 +β Z 1

0

x^αf^β(x)dx,

that is

(2.11) β Z 1

0

f^α+β(x)dx+ α

α+β+ 1 ≥ α

α+β+ 1 +β Z 1

0

x^αf^β(x)dx, which completes this proof.

Lastly, we generalize our result.

Theorem 2.4. Letf(x)≥0be a continuous function on[0, b], b ≥0satisfying (2.12)

Z b

x

f^β(t)dt≥ Z b

x

t^βdt, ∀x∈[0, b].

Then the inequality (2.13)

Z b

0

f^α+β(x)dx≥ Z b

0

x^αf^β(x)dx hold for every positive real numberα >0andβ >0.

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To prove Theorem2.4, we need the following lemma.

Lemma 2.5. Under the conditions of Theorem2.4, we have (2.14)

Z b

0

x^αf^β(x)dx≥ b^α+β+1 α+β+ 1. Proof. Integrating by parts, we have

Z b

0

x^α−1 Z b

x

f^β(t)dt

dx= 1 α

Z b

0

Z b

x

f^β(t)dt

d(x^α) (2.15)

= 1 α

x^α

Z b

x

f^β(t)dt ^x=b

x=0

+ 1 α

Z b

0

x^αf^β(x)dx

= 1 α

Z b

0

x^αf^β(x)dx,

which yields (2.16)

Z b

0

x^αf^β(x)dx=α Z b

0

x^α−1 Z b

x

f^β(t)dt

dx.

On the other hand, by (2.12), we get Z b

0

x^α−1 Z b

x

f^β(t)dt

dx≥ Z b

0

x^α−1 Z b

x

t^βdt

dx (2.17)

= 1

β+ 1 Z b

0

x^α−1(b^β+1−x^β+1)dx

= b^α+β+1 α(α+β+ 1). Therefore, (2.14) holds.

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We now give the proof of Theorem2.4.

Proof of Theorem2.4. Using Lemma2.2, we obtain

(2.18) β

Z b

0

f^α+β(x)dx+α Z b

0

x^α+βdx≥(α+β) Z b

0

x^αf^β(x)dx.

Moreover, by using Lemma2.5, we get (α+β)

Z b

0

x^αf^β(x)dx =α Z b

0

x^αf^β(x)dx+β Z b

0

x^αf^β(x)dx (2.19)

≥α b^α+β+1 α+β+ 1 +β

Z b

0

x^αf^β(x)dx,

that is (2.20) β

Z b

0

f^α+β(x)dx+α b^α+β+1

α+β+ 1 ≥α b^α+β+1 α+β+ 1 +β

Z b

0

x^αf^β(x)dx, which completes the proof.

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References

[1] J.-CH. KUANG, Applied Inequalities, 3rd edition, Shandong Science and Tech- nology Press, Jinan, China, 2004. (Chinese)

[2] Q.A. NGÔ, D.D. THANG, T.T. DAT AND D.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].