Journal of Inequalities in Pure and Applied Mathematics

Transcription

1 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, Việt Nam National University, Hà Nội, Việt Nam anhngq@yahoo.com.vn thangdd@vnu.edu.vn dattt@vnu.edu.vn datuan11@yahoo.com volume 7, issue 4, article 12, 26. Received 3 May, 26; accepted 19 May, 26. Communicated by: P.S. Bullen Abstract Home Page c 2 Victoria University ISSN (electronic):

2 Abstract In this paper, some integral inequalities are presented by analytic approach. An open question will be proposed later on. 2 Mathematics Subject Classification: 26D15. Key words: Integral inequality. 1 Introduction The Case of Natural Numbers The Case of Real Numbers References Page 2 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

3 1. Introduction Let f() be a continuous function on [, 1] satisfying (1.1) f (t) dt 1 2, [, 1]. 2 Firstly, we consider an integral inequality below. Lemma 1.1. If (1.1) holds then we have (1.2) [f ()] 2 d f () d. 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 which yields (f () ) 2 d = f 2 () d 2 f 2 () d 2 f () d 1 3. f () d + 2 d, Page 3 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

4 Let A := ( 1 ). f (t) dt By using our assumption we have ( ) 1 2 A = f (t) dt d = On the other hand, integrating by parts, we also get ( ) A = f (t) dt = = f (t) dt f () d. 1 + Thus f () d 1 3, which gives the conclusion. Remark 1. Condition (1.1) can be rewritten as (1.3) f (t) dt f () d tdt, [, 1]. Throughout this paper, we always assume that function f satisfies (1.1), moreover, we also assume that (1.4) f() for every [, 1]. Page 4 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

5 Lemma 1.2. n+1 f () d 1 for all n N. n+3 Proof. We have which yields n ( On the other hand Therefore ) f (t) dt d = 1 n + 1 n+1 f () d = (n + 1) The proof is completed. n ( = 1 n + 1 n+1 ) f (t) dt d n+1 f () d (n + 1) ( + 1 n + 1 n ( ) f (t) dt d ( n+1) f (t) dt =1 = n+1 f () d, ) f (t) dt d. n 1 2 d. 2 n 1 2 d = 1 2 n + 3. Page 5 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

6 2. The Case of Natural Numbers Theorem 2.1. Assume that (1.1) and (1.4) hold. Then for every n N. f n+1 () d n f () d Proof. By using the Cauchy inequality, we obtain f n+1 () + n n+1 (n + 1) n f (). Thus f n+1 () d + n Moreover, by using Lemma 1.2, we get (n + 1) n f () d = n that is f n+1 () d + which completes this proof. n+1 d (n + 1) n n n n + 2 n f () d + n n n f () d, n f () d. n f () d n f () d, Page 6 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

7 Theorem 2.2. Assume that (1.1) and (1.4) hold. Then for every n N. Proof. It is known that f n+1 () d f n () d that is (f n () n ) (f () ), [, 1], f n+1 () + n+1 n f () + f n (), [, 1]. By integrating with some simple calculation we conclude that f n+1 () d + 1 n + 2 Once again, by Lemma 1.2, we obtain which gives the conclusion. n f () d + f n+1 () d + 1 n n f n () d. f n () d, 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 (2.1) α+1 f () d 1, α >. α + 3 Page 7 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

8 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 and y, we always have α + βy a y β. Theorem 3.2. Assume that (1.1) and (1.4) hold. Then f α+1 () d for every positive real number α >. Proof. Using Theorem 3.1 we get which gives 1 α f () d 1 f α+1 () + α α+1 α f (), f α+1 () d + α α+1 d By the same argument together with (2.1) we obtain 1 f α+1 α () d + () (α + 2) 1 α f () d + α α f () d. α f () d Page 8 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

9 Hence 1 1 The present proof is completed. α f () d + f α+1 () d 1 Theorem 3.3. Assume that (1.1) and (1.4) hold. Then f α+1 () d for every positive real number α >. f α () d α () (α + 2). α f () d. 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() be a continuous function on [, 1] satisfying f (t) dt Under what conditions does the inequality hold for α and β? f α+β () d tdt, [, 1]. t α f β () d. Page 9 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26

10 References [1] E.F. BECKENBACH AND R. BELLMAN, Inequalities, Springer, Berlin, [2] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, 2nd edition, Cambridge University Press, Cambridge, [3] JI-CHANG KUANG, Applied Inequalities, 2nd edition, Hunan Education Press, Changsha, China, [4] D.S. MITRINOVIĆ, Analytic Inequalities, Springer-Verlag, Berlin, 197. [5] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, Page 1 of 1 J. Ineq. Pure and Appl. Math. 7(4) Art. 12, 26