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volume 1, issue 2, article 19, 2000.

Received 27 January, 2000;

accepted 12 April, 2000.

Communicated by:J. Sandor

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Journal of Inequalities in Pure and Applied Mathematics

SEVERAL INTEGRAL INEQUALITIES

FENG QI

Department of Mathematics Jiaozuo Institute of Technology Jiaozuo City, Henan 454000

THE PEOPLE’S REPUBLIC OF CHINA EMail:[email protected]

URL:http://rgmia.vu.edu.au/qi.html

c

2000Victoria University ISSN (electronic): 1443-5756 001-00

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Several Integral Inequalities Feng Qi

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J. Ineq. Pure and Appl. Math. 1(2) Art. 19, 2000

http://jipam.vu.edu.au

Abstract

In the article, some integral inequalities are presented by analytic approach and mathematical induction. An open problem is proposed.

2000 Mathematics Subject Classification:26D15 Key words: Integral inequality, mathematical induction.

The author was supported in part by NSF of Henan Province, SF of the Education Committee of Henan Province (No. 1999110004), and Doctor Fund of Jiaozuo Insti- tute of Technology, The People’s Republic of China

1. Several Integral Inequalities

In this article, we establish some integral inequalities by analytic method and induction.

Proposition 1.1. Let f(x) be differentiable on (a, b) and f(a) = 0. If 0 6 f⁰(x)61, then

(1.1)

Z b

a

f(x)3

dx6 Z b

a

f(x)dx 2

.

If f⁰(x) > 1, then inequality (1.1) reverses. The equality in (1.1) holds only if f(x)≡0orf(x) =x−a.

Proof. Fora6t 6b, set F(t) =

Z t

a

f(x)dx 2

− Z t

a

f(x)3

dx.

Simple computation yields F⁰(t) =

2

Z t

a

f(x)dx−

f(t)2

f(t),G(t)f(t), G⁰(t) = 2

1−f⁰(t) f(t).

Sincef⁰(t)>0andf(a) = 0, thusf(t)is increasing andf(t)>0.

(1) When06f⁰(t)61, we haveG⁰(t)>0,G(t)increases andG(t)>0be- cause ofG(a) = 0, henceF⁰(t) =G(t)f(t)>0,F(t)is increasing. Since F(a) = 0, we have F(t) > 0, and F(b) > 0. Therefore, the inequality (1.1) holds.

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(2) Whenf⁰(t)>1, we haveG⁰(t)60,G(t)decreases,G(t)60,F⁰(t)60, andF(t)is decreasing, thenF(t)60, the inequality (1.1) reverses.

(3) Since the equality in (1.1) holds only iff⁰(t) = 1orf(t) = 0, substitution off(t) =t+cinto (1.1) and standard argument leads toc=−a.

The proof is completed.

Corollary 1.2. [3, p. 624] Let f(x) be a continuous function on the closed interval[0,1]andf(0) = 0, its derivative of the first order is bounded by 06 f⁰(x)61forx∈(0,1). Then

(1.2)

Z 1

0

f(x)3

dx6 Z 1

0

f(x)dx 2

. Equality in (1.2) holds if and only iff(x) = 0orf(x) =x.

Proposition 1.3. Supposef(x)has continuous derivative of then-th order on the interval[a, b],f⁽ⁱ⁾(a)>0andf⁽ⁿ⁾(x)>n!, where06i6n−1, then (1.3)

Z b

a

f(x)n+2

dx>

Z b

a

f(x)dx n+1

. Proof. Let

(1.4) H(t) = Z t

a

f(x)n+2

dx− Z t

a

f(x)dx n+1

, t ∈[a, b].

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Direct calculation produces H⁰(t) =

f(x)n+1

−(n+ 1) Z t

a

f(x)dx n

f(t),h₁(t)f(t), h⁰₁(t) =(n+ 1)

f(x)n−1

f⁰(t)−n Z t

a

f(x)dx ⁿ⁻¹

f(t),(n+ 1)h₂(t)f(t), h⁰₂(t) =

f(x)ⁿ⁻²

f⁰⁰(t) + (n−1)

f(t)ⁿ⁻³ f⁰(t)2

−n(n−1) Z t

a

f(x)dx ⁿ⁻²

f(t),h₃(t)f(t).

By induction, we obtain (1.5)

h⁰_i(t) =

f⁽ⁱ⁾(t)

f(t)ⁿ⁻ⁱ

+pi(t)− n!

(n−i)!

Z t

a

f(x)dx n−i

f(t),hi+1(t)f(t), where26i6nand

p₂(t) = (n−1)

f(t)ⁿ⁻³ f⁰(t)2

, p_i+1(t)f(t) =p⁰_i(t) + (n−i)f⁽ⁱ⁾(t)

f(t)n−i−1

f⁰(t).

(1.6)

From f⁽ⁿ⁾(t) > n! and f⁽ⁱ⁾(a) > 0 for 0 6 i 6 n −1, it follows that f⁽ⁱ⁾(t)>0and are increasing for06i6n−1.

Using mathematical induction, it is easy to see that p_i(t) = X

j0+

i−1

P

k=1

k·j_k=n−1

C(j₀, j₁, . . . , ji−1)

i−1

Y

k=0

f^(k)(t)j_k

,

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wherej_kandC(j₀, j₁, . . . , ji−1)are nonnegative integers,06k6i−1.

Therefore, we obtainp⁰_k(t)> 0andpk+1(t) >0, thenp⁰_k−1(t)andpk(t)are increasing for26k6n. Straightforward computation yields

h_n+1(t) =f⁽ⁿ⁾(t) +p_n(t)−n!.

Considering f⁽ⁿ⁾(t) > n!, we get h_n+1(t) > 0, and h⁰_n(t) > 0, then h_n(t) increases.

By our definitions ofh_i(t), we have, for16i6n−1, h_i+1(a) =f⁽ⁱ⁾(a)

f(a)n−i

+p_i(a)>0.

Therefore, using induction oni, we obtainh⁰_i(t) > 0, h_i(t) > 0, andh_i(t) are increasing for1 6 i 6 n. ThenH⁰(t) > 0and increases, andH(t) > 0. The inequality (1.3) follows fromH(b)>0. Thus, Proposition1.3is proved.

Corollary 1.4. Let f(x) be n-times differentiable on [a, b], f⁽ⁱ⁾(a) > 0 and f⁽ⁿ⁾(x) > n! for 0 6 i 6 n −1. Then the functions H(t), h_j(t) and p_k(t) defined by the formulae (1.4), (1.5) and (1.6) are increasing and convex, where 16j 6n−1and26k 6n−2.

Remark 1.1. The inequality (1.3) is not found in [1,2, 4,5]. So maybe it is a new inequality.

Lastly, we propose the following open problem:

Theorem 1.5 (Open Problem). Under what conditions does the inequality (1.7)

Z b

a

f(x)t

dx>

Z b

a

f(x)dx t−1

hold fort >1?

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References

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

[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. (Chinese)

[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.