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:
EMail:
EMail:
EMail:
anhngq@yahoo.com.vn
thangdd@vnu.edu.vn
dattt@vnu.edu.vn
datuan11@yahoo.com
volume 7, issue 4, article 120,
2006.
Received 03 May, 2006;
accepted 19 May, 2006.
Communicated by: P.S. Bullen
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
130-06
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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.
Notes on an Integral Inequality
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
The Case of Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
3
The Case of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
6
8
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
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1.
Introduction
Let f (x) be a continuous function on [0, 1] satisfying
Z 1
1 − x2
(1.1)
f (t) dt ≥
, ∀x ∈ [0, 1] .
2
x
Firstly, we consider an integral inequality below.
Lemma 1.1. If (1.1) holds then we have
Z 1
Z
2
(1.2)
[f (x)] dx ≥
0
Notes on an Integral Inequality
1
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
xf (x) dx.
0
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 Lemma 1.1 and
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 Lemma 1.1.
Proof of Lemma 1.1. It is known that
Z 1
Z 1
Z
2
2
0≤
(f (x) − x) dx =
f (x) dx − 2
0
0
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1
Z
xf (x) dx +
0
0
1
x2 dx,
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which yields
Z
1
2
Z
f (x) dx ≥ 2
0
0
1
1
xf (x) dx − .
3
J. Ineq. Pure and Appl. Math. 7(4) Art. 120, 2006
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Let A :=
R 1 R 1
0
f
(t)
dt
. By using our assumption we have
x
Z 1
Z 1 Z 1
1 − x2
1
A=
f (t) dt ≥
dx = .
2
3
0
x
0
On the other hand, integrating by parts, we also get
Z 1 Z 1
A=
f (t) dt
0
x
1
Z
=x
Z
=
x
1 Z
f (t) dt +
0
1
Notes on an Integral Inequality
xf (x) dx
0
1
xf (x) dx.
0
Title Page
Thus
Z
0
1
1
xf (x) dx ≥ ,
3
which gives the conclusion.
Remark 1. Condition (1.1) can be rewritten as
Z 1
Z 1
(1.3)
f (t) dt ≥
tdt, ∀x ∈ [0, 1] .
x
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Throughout this paper, we always assume that function f satisfies (1.1),
moreover, we also assume that
(1.4)
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
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f (x) ≥ 0
J. Ineq. Pure and Appl. Math. 7(4) Art. 120, 2006
for every x ∈ [0, 1].
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R1
Lemma 1.2.
0
xn+1 f (x) dx ≥
Proof. We have
Z 1 Z
xn
0
1
1
n+3
f (t) dt dx =
x
for all n ∈ N.
1
Z
1
n+1
1
Z
0
f (t) dt d xn+1
x
1
x=1
f (t) dt
x
x=0
Z 1
1
+
xn+1 f (x) dx,
n+1 0
1
=
xn+1
n+1
which yields
Z
1
x
n+1
1
Z
f (x) dx = (n + 1)
x
0
n
Z
0
On the other hand
Z
1
x
n
Z
0
1
1
Z
f (t) dt dx.
x
Z
f (t) dt dx ≥
x
1
0
xn
1 − x2
dx.
2
Notes on an Integral Inequality
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
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Therefore
Z
1
x
n+1
Z
f (x) dx ≥ (n + 1)
0
The proof is completed.
1
x
0
n1
− x2
1
dx =
.
2
n+3
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J. Ineq. Pure and Appl. Math. 7(4) Art. 120, 2006
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2.
The Case of Natural Numbers
Theorem 2.1. Assume that (1.1) and (1.4) hold. Then
Z 1
Z 1
n+1
f
(x) dx ≥
xn f (x) dx
0
0
for every n ∈ N.
Proof. By using the Cauchy inequality, we obtain
Notes on an Integral Inequality
f n+1 (x) + nxn+1 ≥ (n + 1) xn f (x) .
Thus
Z
1
f
n+1
Z
(x) dx + n
0
1
x
n+1
Z
dx ≥ (n + 1)
0
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
1
xn f (x) dx.
0
Moreover, by using Lemma 1.2, we get
Z 1
Z 1
Z 1
n
n
(n + 1)
x f (x) dx = n
x f (x) dx +
xn f (x) dx
0
0
0
Z 1
n
≥
+
xn f (x) dx,
n+2
0
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that is
Z
1
f n+1 (x) dx +
0
which completes this proof.
n
n
≥
+
n+2
n+2
Z
1
xn f (x) dx,
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Theorem 2.2. Assume that (1.1) and (1.4) hold. Then
Z 1
Z 1
n+1
f
(x) dx ≥
xf n (x) dx
0
0
for every n ∈ N.
Proof. It is known that
(f n (x) − xn ) (f (x) − x) ≥ 0,
∀x ∈ [0, 1] ,
Notes on an Integral Inequality
that is
f n+1 (x) + xn+1 ≥ xn f (x) + xf n (x) ,
∀x ∈ [0, 1] .
By integrating with some simple calculation we conclude that
Z 1
Z 1
Z 1
1
n
n+1
f
(x) dx +
≥
x f (x) dx +
xf n (x) dx.
n
+
2
0
0
0
Once again, by Lemma 1.2, we obtain
Z 1
Z 1
1
1
n+1
f
(x) dx +
≥
+
xf n (x) dx,
n+2
n+2
0
0
which gives the conclusion.
Remark 2. By the same argument, we see that the result of Lemma 1.2 also
holds when n is a positive real number. That is
Z 1
1
(2.1)
xα+1 f (x) dx ≥
, ∀α > 0.
α+3
0
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
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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). Let α and β be positive real numbers satisfying α + β = 1. Then for every positive real numbers x and y, we
always have
αx + βy ≥ xa y β .
Theorem 3.2. Assume that (1.1) and (1.4) hold. Then
Z 1
Z 1
α+1
f
(x) dx ≥
xα f (x) dx
0
Notes on an Integral Inequality
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
0
for every positive real number α > 0.
Proof. Using Theorem 3.1 we get
1
α
f α+1 (x) +
xα+1 ≥ xα f (x) ,
α+1
α+1
which gives
Z 1
Z 1
Z 1
1
α
α+1
α+1
f
(x) dx +
x dx ≥
xα f (x) dx.
α+1 0
α+1 0
0
By the same argument together with (2.1) we obtain
Z 1
1
α
f α+1 (x) dx +
α+1 0
(α + 1) (α + 2)
Z 1
Z 1
1
α
α
≥
x f (x) dx +
xα f (x) dx
α+1 0
α+1 0
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1
≥
α+1
Z
1
xα f (x) dx +
0
α
.
(α + 1) (α + 2)
Hence
Z 1
Z 1
1
1
α+1
f
(x) dx ≥
xα f (x) dx.
α+1 0
α+1 0
The present proof is completed.
Theorem 3.3. Assume that (1.1) and (1.4) hold. Then
Z 1
Z 1
α+1
f
(x) dx ≥
xf α (x) dx
0
0
Notes on an Integral Inequality
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
for every positive real number α > 0.
The proof of Theorem 3.3 is similar to the proof of Theorem 2.2 therefore,
we omit it.
Lastly, we propose the following open problem.
Open Problem. Let f (x) be a continuous function on [0, 1] satisfying
Z 1
Z 1
f (t) dt ≥
tdt, ∀x ∈ [0, 1] .
x
x
hold for α and β?
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Under what conditions does the inequality
Z
Z 1
α+β
f
(x) dx ≥
0
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1
tα f β (x) dx.
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References
[1] E.F. BECKENBACH
1983.
AND
R. BELLMAN, Inequalities, Springer, Berlin,
[2] G.H. HARDY, J.E. LITTLEWOOD AND G. 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. MITRINOVIĆ, Analytic Inequalities, Springer-Verlag, Berlin, 1970.
[5] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993.
Notes on an Integral Inequality
Quô´c Anh Ngô, Du Duc Thang,
Tran Tat Dat and Dang Anh Tuan
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