JJ II

(1)

volume 7, issue 4, article 120, 2006.

Received 03 May, 2006;

accepted 19 May, 2006.

Communicated by:P.S. Bullen

Abstract Contents

JJ II

J I

Home Page Go Back

Close Quit

Journal of Inequalities in Pure and Applied Mathematics

NOTES ON AN INTEGRAL INEQUALITY

QUÔ´C ANH NGÔ, DU DUC THANG, TRAN TAT DAT, AND DANG ANH TUAN

Department of Mathematics, Mechanics and Informatics, College of Science, Viê.t Nam National University, Hà Nô.i, Viê.t Nam

EMail:anhngq@yahoo.com.vn EMail:thangdd@vnu.edu.vn EMail:dattt@vnu.edu.vn EMail:datuan11@yahoo.com

c

2000Victoria University ISSN (electronic): 1443-5756 130-06

(2)

Notes on an Integral Inequality

Quô´c Anh Ngô, Du Duc Thang, Tran Tat Dat and Dang Anh Tuan

Title Page Contents

JJ II

J I

Go Back Close

Quit Page2of10

J. Ineq. Pure and Appl. Math. 7(4) Art. 120, 2006

http://jipam.vu.edu.au

Abstract

In this paper, some integral inequalities are presented by analytic approach. An open question will be proposed later on.

2000 Mathematics Subject Classification:26D15.

Key words: Integral inequality.

1. Introduction

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

(1.1)

Z 1

x

f(t)dt ≥ 1−x²

2 , ∀x∈[0,1].

Firstly, we consider an integral inequality below.

Lemma 1.1. If (1.1) holds then we have (1.2)

Z 1

0

[f(x)]²dx≥ Z 1

0

xf(x)dx.

The aim of this paper is to generalize (1.2) in order to obtain some new integral inequalities. In the first part of this paper, we will prove Lemma1.1and present some preliminary results. Our main results are Theorem 2.1, Theorem 2.2 which will be proved in Section 2 and Theorem 3.2, Theorem 3.3 which will be proved in Section 3. Finally, an open question is proposed. And now, we begin with a proof of Lemma1.1.

Proof of Lemma1.1. It is known that

0≤ Z 1

0

(f(x)−x)²dx= Z 1

0

f²(x)dx−2 Z 1

0

xf(x)dx+ Z 1

0

x²dx,

which yields

Z 1

0

f²(x)dx≥2 Z 1

0

xf(x)dx− 1 3.

(4)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page4of10

LetA:=R1 0

R1

x f(t)dt

. By using our assumption we have

A= Z 1

0

Z 1

x

f(t)dt

≥ Z 1

0

1−x²

2 dx= 1 3. On the other hand, integrating by parts, we also get

A= Z 1

0

Z 1

x

f(t)dt

=x Z 1

x

f(t)dt

1

0

+ Z 1

0

xf(x)dx

= Z 1

0

xf(x)dx.

Thus

Z 1

0

xf(x)dx≥ 1 3, which gives the conclusion.

Remark 1. Condition (1.1) can be rewritten as (1.3)

Z 1

x

f(t)dt ≥ Z 1

x

tdt, ∀x∈[0,1].

Throughout this paper, we always assume that function f satisfies (1.1), moreover, we also assume that

(1.4) f(x)≥0

for everyx∈[0,1].

(5)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page5of10

Lemma 1.2. R1

0 xⁿ⁺¹f(x)dx≥ _n+3¹ for alln∈N. Proof. We have

Z 1

0

xⁿ Z 1

x

f(t)dt

dx= 1 n+ 1

Z 1

0

Z 1

x

f(t)dt

d xⁿ⁺¹

= 1

n+ 1xⁿ⁺¹ Z 1

x

f(t)dt

x=1

x=0

+ 1

n+ 1 Z 1

0

xⁿ⁺¹f(x)dx,

which yields Z 1

0

xⁿ⁺¹f(x)dx= (n+ 1) Z 1

0

xⁿ Z 1

x

f(t)dt

dx.

On the other hand Z 1

0

xⁿ Z 1

x

f(t)dt

dx≥ Z 1

0

xⁿ1−x² 2 dx.

Therefore Z 1

0

xⁿ⁺¹f(x)dx≥(n+ 1) Z 1

0

xⁿ1−x²

2 dx = 1 n+ 3. The proof is completed.

(6)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page6of10

2. The Case of Natural Numbers

Theorem 2.1. Assume that (1.1) and (1.4) hold. Then Z 1

0

fⁿ⁺¹(x)dx≥ Z 1

0

xⁿf(x)dx

for everyn∈N.

Proof. By using the Cauchy inequality, we obtain

fⁿ⁺¹(x) +nxⁿ⁺¹ ≥(n+ 1)xⁿf(x). Thus

Z 1

0

fⁿ⁺¹(x)dx+n Z 1

0

xⁿ⁺¹dx≥(n+ 1) Z 1

0

xⁿf(x)dx.

Moreover, by using Lemma1.2, we get

(n+ 1) Z 1

0

xⁿf(x)dx=n Z 1

0

xⁿf(x)dx+ Z 1

0

xⁿf(x)dx

≥ n n+ 2 +

Z 1

0

xⁿf(x)dx,

that is

Z 1

0

fⁿ⁺¹(x)dx+ n

n+ 2 ≥ n n+ 2 +

Z 1

0

xⁿf(x)dx, which completes this proof.

(7)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page7of10

0

fⁿ⁺¹(x)dx≥ Z 1

0

xfⁿ(x)dx

for everyn∈N. Proof. It is known that

(fⁿ(x)−xⁿ) (f(x)−x)≥0, ∀x∈[0,1], that is

fⁿ⁺¹(x) +xⁿ⁺¹ ≥xⁿf(x) +xfⁿ(x), ∀x∈[0,1]. By integrating with some simple calculation we conclude that

Z 1

0

fⁿ⁺¹(x)dx+ 1 n+ 2 ≥

Z 1

0

xⁿf(x)dx+ Z 1

0

xfⁿ(x)dx.

Once again, by Lemma1.2, we obtain Z 1

0

fⁿ⁺¹(x)dx+ 1

n+ 2 ≥ 1 n+ 2 +

Z 1

0

xfⁿ(x)dx,

which gives the conclusion.

Remark 2. By the same argument, we see that the result of Lemma 1.2 also holds whennis a positive real number. That is

(2.1)

Z 1

0

x^α+1f(x)dx≥ 1

α+ 3, ∀α >0.

(8)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page8of10

3. The Case of Real Numbers

In order to generalize our results, the case of positive real numbers, we recall another version of the Cauchy inequality as follows.

Theorem 3.1 (General Cauchy inequality). Letαandβbe positive real num- bers satisfying α+β = 1. Then for every positive real numbersx andy, we always have

αx+βy ≥x^ay^β. Theorem 3.2. Assume that (1.1) and (1.4) hold. Then

Z 1

0

f^α+1(x)dx≥ Z 1

0

x^αf(x)dx

for every positive real numberα >0.

Proof. Using Theorem3.1we get 1

α+ 1f^α+1(x) + α

α+ 1x^α+1 ≥x^αf(x), which gives

1 α+ 1

Z 1

0

f^α+1(x)dx+ α α+ 1

Z 1

0

x^α+1dx≥ Z 1

0

x^αf(x)dx.

By the same argument together with (2.1) we obtain 1

α+ 1 Z 1

0

f^α+1(x)dx+ α

(α+ 1) (α+ 2)

≥ 1 α+ 1

Z 1

0

x^αf(x)dx+ α α+ 1

Z 1

0

x^αf(x)dx

(9)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page9of10

≥ 1 α+ 1

Z 1

0

x^αf(x)dx+ α

(α+ 1) (α+ 2). Hence

1 α+ 1

Z 1

0

f^α+1(x)dx≥ 1 α+ 1

Z 1

0

x^αf(x)dx.

The present proof is completed.

0

f^α+1(x)dx≥ Z 1

0

xf^α(x)dx

for every positive real numberα >0.

The proof of Theorem3.3 is similar to the proof of Theorem2.2therefore, we omit it.

Lastly, we propose the following open problem.

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

x

f(t)dt ≥ Z 1

x

tdt, ∀x∈[0,1].

Under what conditions does the inequality Z 1

0

f^α+β(x)dx≥ Z 1

0

t^αf^β(x)dx.

hold forαandβ?

(10)

Title Page Contents

JJ II

J I

Go Back Close

Quit Page10of10

References

[1] E.F. BECKENBACH AND R. BELLMAN, Inequalities, Springer, Berlin, 1983.

[2] G.H. HARDY, J.E. LITTLEWOODANDG. PÓLYA, Inequalities, 2nd edition, Cambridge University Press, Cambridge, 1952.

[3] JI-CHANG KUANG, Applied Inequalities, 2nd edition, Hunan Education Press, Changsha, China, 1993.

[4] D.S. MITRINOVI ´C, Analytic Inequalities, Springer-Verlag, Berlin, 1970.

[5] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993.