Volume 8 (2007), Issue 3, Article 77, 3 pp.
ON AN OPEN QUESTION REGARDING AN INTEGRAL INEQUALITY
K. BOUKERRIOUA AND A. GUEZANE-LAKOUD DEPARTMENT OFMATHEMATICS
UNIVERSITY OFGUELMA
GUELMA, ALGERIA
[email protected] UNIVERSITYBADJIMOKHTAR, ANNABA
ANNABA, ALGERIA
Received 16 January, 2007; accepted 14 July, 2007 Communicated by J.E. Peˇcari´c
ABSTRACT. In the paper "Notes on an integral inequality" published in J. Inequal. Pure &
Appl. Math., 7(4) (2006), Art. 120, an open question was posed. In this short paper, we give the solution and we generalize the results of the mentioned paper.
Key words and phrases: Integral inequalitiy, AG inequality.
2000 Mathematics Subject Classification. 26D15.
1. INTRODUCTION
The following open question was proposed in the paper [1]:
Under what conditions does the inequality (1.1)
Z 1
0
fα+β(x)dx≥ Z 1
0
xβfα(x)dx hold forαandβ?
In the above paper, the authors established some integral inequalities and derived their results using an analytic approach.
In the present paper, we give a solution and further generalization of the integral inequalities presented in [1].
2. THEANSWER TO THEPOSEDQUESTION
Throughout this paper, we suppose that f(x)is a continuous and nonnegative function on [0,1].
In [1]], the following lemma was proved.
The authors thank the referee for making several suggestions for improving the presentation of this paper.
028-07
2 K. BOUKERRIOUA ANDA. GUEZANE-LAKOUD
Lemma 2.1. Iff satisfies (2.1)
Z 1
x
f(t)dt≥ 1−x2
2 , ∀x∈[0,1], then
(2.2)
Z 1
0
xα+1f(x)dx≥ 1
α+ 3, ∀α >0.
Theorem 2.2. If the functionf satisfies (2.1), then the inequality (2.3)
Z 1
0
xβfα(x)dx≥ 1 α+β+ 1 holds for every realα≥1andβ >0.
Proof. Applying the AG inequality, we get
(2.4) 1
αfα(x) + α−1
α xα ≥f(x)xα−1.
Multiplying both sides of(2.4)by xβ and integrating the resultant inequality from 0 to 1, we obtain
(2.5)
Z 1
0
xβfα(x)dx+ α−1
α+β+ 1 ≥α Z 1
0
xα+β−1f(x)dx.
Taking into account Lemma 2.1, we have Z 1
0
xβfα(x)dx+ α−1
α+β+ 1 ≥ α α+β+ 1. That is,
Z 1
0
xβfα(x)dx≥ 1 α+β+ 1.
This completes the proof.
Theorem 2.3. If the functionf satisfies (2.1), then (2.6)
Z 1
0
fα+β(x)dx≥ Z 1
0
xβfα(x)dx
for every real α≥1andβ >0.
Proof. Using the AG inequality, we obtain
(2.7) α
α+βfα+β(x) + β
α+βxα+β ≥xβfα(x). Integrating both sides of(2.7), we get
(2.8) α
α+β Z 1
0
fα+β(x)dx+ β
(α+β)(α+β+ 1) ≥ Z 1
0
xβfα(x)dx.
From
Z 1
0
xβfα(x)dx= α α+β
Z 1
0
xβfα(x)dx+ β α+β
Z 1
0
xβfα(x)dx
and by virtue of Theorem 2.3, it follows that (2.9)
Z 1
0
xβfα(x)dx≥ α α+β
Z 1
0
xβfα(x)dx+ β
(α+β)(α+β+ 1).
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 77, 3 pp. http://jipam.vu.edu.au/
ON AN OPEN QUESTION REGARDING AN INTEGRAL INEQUALITY 3
From this inequality and using(2.8)we have, α
α+β Z 1
0
fα+β(x)dx≥ α α+β
Z 1
0
xβfα(x)dx.
Thus (2.6) is proved.
REFERENCES
[1] Q.A. NGÔ, D.D. THANG, T.T. DAT AND D.A. TUAN, Notes On an integral inequality, J. In- equal. Pure & Appl. Math., 7(4) (2006), Art. 120. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=737].
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 77, 3 pp. http://jipam.vu.edu.au/
