Several Integral Inequalities
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Qi, Feng (1999) Several Integral Inequalities. RGMIA research report collection, 2 (7).
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SEVERAL INTEGRAL INEQUALITIES
FENG QI
Abstract. In the article, some integral inequalities are presented by analytic approach and mathematical induction. An open problem is proposed.
In this article, we establish some integral inequalities by analytic method and induction.
Proposition 1. Let f(x) be differentiable on (a, b) and f(a) = 0. If 0 6 f0(x)61, then
Z b
a
f(x)3
dx6 Z b
a
f(x) dx 2
. (1)
If f0(x) >1, then inequality (1) reverses. The equality in (1) holds only if f(x)≡0 or f(x) =x−a.
Proof. Fora6t6b, 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).
Since f0(t)>0 and f(a) = 0, thus f(t) is increasing andf(t)>0.
(1) When 06 f0(t) 61, we have G0(t) >0, G(t) increases andG(t) >0 because of G(a) = 0, hence F0(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) holds.
(2) Whenf0(t)>1, we haveG0(t)60,G(t) decreases,G(t)60,F0(t)60, and F(t) is decreasing, thenF(t)60, the inequality (1) reverses.
(3) Since the equality in (1) holds only iff0(t) = 1 orf(t) = 0, substitution of f(t) =t+cinto (1) and standard argument leads to c=−a.
1991Mathematics Subject Classification. Primary 26D15.
Key words and phrases. 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 Institute of Technology, The People’s Republic of China.
This paper is typeset usingAMS-LATEX.
1
2 FENG QI
The proof is completed.
Corollary 1 ([3, p. 624]). Let f(x) be a continuous function on the closed interval [0,1] and f(0) = 0, its derivative of the first order is bounded by 06f0(x)61 for x∈(0,1). Then
Z 1 0
f(x)3
dx6 Z 1
0
f(x) dx 2
. (2)
Equality in (2)holds if and only if f(x) = 0 or f(x) =x.
Proposition 2. Suppose f(x) has continuous derivative of the n-th order on the interval[a, b], f(i)(a)>0andf(n)(x)>n!, where06i6n−1, then
Z b a
f(x)n+2
dx>
Z b a
f(x) dx n+1
. (3)
Proof. Let H(t) =
Z t a
f(x)n+2
dx− Z t
a
f(x) dx n+1
, t∈[a, b].
(4)
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
h0i(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), (5)
where 26i6nand
p2(t) = (n−1)
f(t)n−3 f0(t)2
, pi+1(t)f(t) =p0i(t) + (n−i)f(i)(t)
f(t)n−i−1 f0(t).
(6)
¿From f(n)(t) > n! and f(i)(a) > 0 for 0 6 i 6 n−1, it follows that f(i)(t)>0 and are increasing for 06i6n−1.
Using the mathematical induction, it is easy to see that pi(t) = X
j0+i−1P
k=1
k·jk=n−1
C(j0, j1, . . . , ji−1)
i−1
Y
k=0
f(k)(t)jk
,
SEVERAL INTEGRAL INEQUALITIES 3
wherejk and C(j0, j1, . . . , ji−1) are nonnegative integers, 06k6i−1.
Therefore, we obtain p0k(t) > 0 and pk+1(t) >0, then p0k−1(t) and pk(t) are increasing for 26k6n. Straightforward computation yields
hn+1(t) =f(n)(t) +pn(t)−n!.
Considering f(n)(t) > n!, we get hn+1(t) > 0, and h0n(t) > 0, then hn(t) increases.
By definitions ofhi(t), we have, for 16i6n−1, hi+1(a) =f(i)(a)
f(a)n−i
+pi(a)>0.
Therefore, using induction on i, we obtain h0i(t) > 0, hi(t) > 0, and hi(t) are increasing for 16i6n. Then H0(t) >0 and increases, and H(t)>0.
The inequality (3) follows from H(b)>0. The proposition 2 is proved.
Corollary 2. Let f(x) be n-times differentiable on [a, b], f(i)(a) > 0 and f(n)(x) > n! for 0 6 i 6n−1. Then the functions H(t), hj(t) and pk(t) defined by the formulae (4), (5) and (6) are increasing and convex, where 16j6n−1 and 26k6n−2.
Remark. The inequality (3) is not found in [1, 2, 4, 5]. So maybe it is a new inequality.
At last, we propose the following open problem in a natural way:
Open Problem. Under what conditions does the inequality Z b
a
f(x)t
dx>
Z b a
f(x) dx t−1 (7)
hold for t >1 ?
References
[1] E. F. Beckenbach and R. Bellman,Inequalities, Springer, Berlin, 1983.
[2] G. H. Hardy, J. E. Littlewood and G. P´olya, 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. Peˇcari´c and A. M. Fink, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
Department of Mathematics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, The People’s Republic of China
E-mail address: [email protected]
