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:lwjboy@126.com lichunchengcxy@126.com
JIAN-WEI DONG
Department of Mathematics and Physics
Zhengzhou Institute of Aeronautical Industry Management Zhengzhou 450015, China
EMail:dongjianweiccm@163.com
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.
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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Contents
1 Introduction 3
2 Main Results and Proofs 5
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
and Jian-wei Dong
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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
fα+1(x)dx≥ Z 1
0
xαf(x)dx and
(1.3)
Z 1
0
fα+1(x)dx≥ Z 1
0
xfα(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
fα+β(x)dx≥ Z 1
0
xαfβ(x)dx, hold forαandβ?
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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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
fα+β(x)dx≥ Z b
0
xαfβ(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.
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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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
fβ(t)dt≥ Z 1
x
tβdt, ∀x∈[0,1].
Then the inequality (2.2)
Z 1
0
fα+β(x)dx≥ Z 1
0
xαfβ(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 ≥xαyβ.
Lemma 2.3. Under the conditions of Theorem2.1, we have (2.4)
Z 1
0
xαfβ(x)dx≥ 1 α+β+ 1.
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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Proof. Integrating by parts, we have Z 1
0
xα−1 Z 1
x
fβ(t)dt
dx= 1 α
Z 1
0
Z 1
x
fβ(t)dt
d(xα) (2.5)
= 1 α
xα
Z 1
x
fβ(t)dt x=1
x=0
+ 1 α
Z 1
0
xαfβ(x)dx
= 1 α
Z 1
0
xαfβ(x)dx,
which yields (2.6)
Z 1
0
xαfβ(x)dx=α Z 1
0
xα−1 Z 1
x
fβ(t)dt
dx.
On the other hand, by (2.1), we get Z 1
0
xα−1 Z 1
x
fβ(t)dt
dx≥ Z 1
0
xα−1 Z 1
x
tβdt
dx (2.7)
= 1
β+ 1 Z 1
0
(xα−1−xα+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+βfα+β(x) + α
α+βxα+β ≥xαfβ(x),
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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which gives
(2.9) β
Z 1
0
fα+β(x)dx+α Z 1
0
xα+βdx≥(α+β) Z 1
0
xαfβ(x)dx.
Moreover, by using Lemma2.3, we get (α+β)
Z 1
0
xαfβ(x)dx=α Z 1
0
xαfβ(x)dx+β Z 1
0
xαfβ(x)dx (2.10)
≥ α
α+β+ 1 +β Z 1
0
xαfβ(x)dx,
that is
(2.11) β Z 1
0
fα+β(x)dx+ α
α+β+ 1 ≥ α
α+β+ 1 +β Z 1
0
xαfβ(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
fβ(t)dt≥ Z b
x
tβdt, ∀x∈[0, b].
Then the inequality (2.13)
Z b
0
fα+β(x)dx≥ Z b
0
xαfβ(x)dx hold for every positive real numberα >0andβ >0.
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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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
xαfβ(x)dx≥ bα+β+1 α+β+ 1. Proof. Integrating by parts, we have
Z b
0
xα−1 Z b
x
fβ(t)dt
dx= 1 α
Z b
0
Z b
x
fβ(t)dt
d(xα) (2.15)
= 1 α
xα
Z b
x
fβ(t)dt x=b
x=0
+ 1 α
Z b
0
xαfβ(x)dx
= 1 α
Z b
0
xαfβ(x)dx,
which yields (2.16)
Z b
0
xαfβ(x)dx=α Z b
0
xα−1 Z b
x
fβ(t)dt
dx.
On the other hand, by (2.12), we get Z b
0
xα−1 Z b
x
fβ(t)dt
dx≥ Z b
0
xα−1 Z b
x
tβdt
dx (2.17)
= 1
β+ 1 Z b
0
xα−1(bβ+1−xβ+1)dx
= bα+β+1 α(α+β+ 1). Therefore, (2.14) holds.
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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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
fα+β(x)dx+α Z b
0
xα+βdx≥(α+β) Z b
0
xαfβ(x)dx.
Moreover, by using Lemma2.5, we get (α+β)
Z b
0
xαfβ(x)dx =α Z b
0
xαfβ(x)dx+β Z b
0
xαfβ(x)dx (2.19)
≥α bα+β+1 α+β+ 1 +β
Z b
0
xαfβ(x)dx,
that is (2.20) β
Z b
0
fα+β(x)dx+α bα+β+1
α+β+ 1 ≥α bα+β+1 α+β+ 1 +β
Z b
0
xαfβ(x)dx, which completes the proof.
Open Problem Concerning an Integral Inequality Wen-jun Liu, Chun-Cheng Li
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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].