INROADS
Real Analysis Exchange Vol. 33(2), 2007/2008, pp. 471–474
Quˆ o´c-Anh Ngˆ o, Department of Mathematics, College of Science, Viˆe.t Nam National University, H` a Nˆ o.i, Viˆe.t Nam. email: bookworm
[email protected] Feng Qi, Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China. email:
[email protected],
[email protected],
[email protected] Ninh Van Thu, Department of Mathematics, College of Science, Viˆe.t Nam National University, H` a Nˆ o.i, Viˆe.t Nam. email:
[email protected]
NEW GENERALIZATIONS OF AN INTEGRAL INEQUALITY Abstract In this short paper an integral inequality posed in the 11th International Mathematical Competition for University Students is further generalized.
1
Introduction.
Problem 2 of the 11th International Mathematical Competition for University Students, Skopje, Macedonia, 25–26 July 2004 (see [1]) reads as follows: Proposition 1. Let f, g : [a, b] → [0, ∞) be two continuous and non-decreasing functions such that Z xp Z xp f (t) dt ≤ g(t) dt (1) a
a
for x ∈ [a, b] and Z
b
p
Z f (t) dt =
a
b
p g(t) dt.
(2)
a
Key Words: integral inequality, generalization, the second mean value theorem for integrals Mathematical Reviews subject classification: 26D15 Received by the editors October 27, 2007 Communicated by: Alexander Olevskii
471
ˆ´c-Anh Ngo ˆ , Feng Qi, and Ninh Van Thu Quo
472
Then Z
b
p
b
Z
p 1 + g(t) dt.
1 + f (t) dt ≥
a
a
Rbp Considering (2), it is clear that (1) can be rewritten as x f (t) dt ≥ p p Rbp g(t) dt. By replacing f (x) with f (x) and g(x) with g(x), Proposix tion 1 can be simplified into the following Proposition 2. Proposition 2. Let f, g : [a, b] → [0, ∞) be two continuous, non-decreasing functions such that Z b Z b f (t) dt ≥ g(t) dt (3) x
x
for x ∈ [a, b] and Z
b
b
Z f (t) dt =
a
g(t) dt.
(4)
p 1 + g 2 (t) dt.
(5)
a
Then Z
b
p
1+
f 2 (t) dt
b
Z ≥ a
a
Rx Rx Let F (x) = a f (t) dt and G(x) = a g(t) dt. Then F (x) ≤ G(x) for x ∈ [a, b] and G(x) is a convex function on [a, b]. On the other hand, since F (a) = G(a) and F (b) = G(b), then inequality (5) is apparently valid because the length of the curve y = F (x) is not less than that of the curve y = G(x). This explains the geometric meaning of Proposition 2 and gives a solution to Proposition 1. In [3], Proposition 1 and Proposition 2 were generalized as follows: Theorem A. Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) a continuous, non-decreasing function satisfying (3) and (4). Then Z b Z b h(f (t)) dt ≥ h(g(t)) dt (6) x
x
for every convex function h on [0, ∞). Theorem B. Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) a continuous, non-increasing function satisfying the reversed inequalities of (3) and (4). Then inequality (6) holds true for every convex function h on [0, ∞).
473
New Generalizations of an Integral Inequality
The main aim of this paper is to further generalize Proposition 1 and Proposition 2. Our main results are included in the two theorems below. Theorem 1. Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) a continuous, non-decreasing function satisfying (3). Then inequality (6) is valid for every convex function h such that h0 ≥ 0 and h0 is integrable on [0, ∞). Corollary 1. Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → Rb [0, ∞) a continuous, non-decreasing function satisfying (3). Then a f α (t) dt ≥ Rb α g (t) dt for every α > 1. a Theorem 2. Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) a continuous, non-increasing function satisfying the reversed inequality of (3). Then inequality (6) holds true for every convex function h such that h0 ≤ 0 and h0 is integrable on [0, ∞).
2
Proofs of Theorem 1 and Theorem 2.
In order to prove our theorems, the well known second mean value theorem for integrals will be available. Lemma 1 ([2, p. 35]). Let f (x) be bounded and monotonic and let g(x) be integrable on [a, b]. Then there exists some ξ ∈ [a, b] such that Z
b
Z f (x)g(x) dx = f (a)
a
ξ
Z g(x) dx + f (b)
a
b
g(x) dx. ξ
Rb Rb Proof of Theorem 1. Let φ(x) = − x f (t) dt and ϕ(x) = − x g(t) dt. Since h is convex, h(t) ≥ h(s) + (t − s)h0 (s) for a ≤ s, t ≤ b. Therefore h(φ0 (t)) ≥ h(ϕ0 (t)) + [φ0 (t) − ϕ0 (t)]h0 (ϕ0 (t)), which gives Z a
b
h(φ0 (t)) dt ≥
Z a
b
h(ϕ0 (t)) dt +
Z
b
[φ0 (t) − ϕ0 (t)]h0 (ϕ0 (t)) dt.
a
Rb Now it is sufficient to prove that a [φ0 (t) − ϕ0 (t)]h0 (ϕ0 (t)) dt ≥ 0. Since h(t) is convex, the function h0 (t) is non-decreasing. Since g(t) is non-decreasing, the function ϕ0 (t) is also non-decreasing. Thus the composite function h0 (ϕ0 (t)) is
ˆ´c-Anh Ngo ˆ , Feng Qi, and Ninh Van Thu Quo
474
non-decreasing with respect to t. Using Lemma 1 for some ξ ∈ [a, b] yields b
Z
[φ0 (t) − ϕ0 (t)]h0 (ϕ0 (t)) dt
a 0
0
ξ
Z
0
0
0
0
Z
[φ (t) − ϕ (t)] dt + h (ϕ (b))
= h (ϕ (a)) a
b
[φ0 (t) − ϕ0 (t)] dt
ξ
= h0 (g(a))[φ(ξ) − φ(a) − ϕ(ξ) + ϕ(a)] + h0 (g(b))[φ(b) − φ(ξ) − ϕ(b) + ϕ(ξ)] ≥ h0 (g(a))[φ(ξ) − ϕ(ξ)] + h0 (g(b))[φ(b) − φ(ξ) − ϕ(b) + ϕ(ξ)] = [φ(ξ) − ϕ(ξ)][h0 (g(a)) − h0 (g(b))] ≥ 0, since φ(ξ) ≤ ϕ(ξ) and 0 ≤ h0 (g(a)) ≤ h0 (g(b)). The proof of Theorem 1 is complete. Proof of Theorem 2. Considering the proof of Theorem 1, it suffices to Rb prove a [φ0 (t) − ϕ0 (t)][−h0 (ϕ0 (t))] dt ≤ 0. Since h is convex, −h0 is nonincreasing. Since g is non-increasing, ϕ0 is also non-increasing. Consequently, the composite function −h0 (ϕ0 (t)) is non-increasing with respect to t. Utilizing Lemma 1 for some ξ ∈ [a, b] leads to Z
b
[φ0 (t) − ϕ0 (t)]h0 (ϕ0 (t)) dt
a
= h0 (ϕ0 (a))
Z
ξ
[φ0 (t) − ϕ0 (t)] dt + h0 (ϕ0 (b))
a
Z
b
[φ0 (t) − ϕ0 (t)] dt
ξ
= h0 (ϕ0 (a))(φ(ξ) − φ(a) − ϕ(ξ) + ϕ(a)) + h0 (ϕ0 (b))(φ(b) − φ(ξ) − ϕ(b) + ϕ(ξ)) ≥ [φ(ξ) − ϕ(ξ)][h0 (g(a)) − h0 (g(b))] ≥ 0, since φ(ξ) ≥ ϕ(ξ) and 0 ≥ h0 (g(a)) ≥ h0 (g(b). The proof is complete.
References [1] The 11th International Mathematical Competition for University Students, Skopje, Macedonia, 25–26 July 2004, Solutions for problems on Day 2, Available online at http://www.imc-math.org.uk/index.php?year= 2004 or http://www.imc-math.org.uk/imc2004/day2_solutions.pdf. [2] M. J. Cloud & B. C. Drachman, Inequalities with Applications to Engineering, Springer, 1998. [3] Q.-A. Ngˆ o & F. Qi, Generalizations of an integral inequality, preprint.
