Generalizations of an Integral Inequality
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Ngô, Quốc Anh and Qi, Feng (2008) Generalizations of an Integral Inequality.
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GENERALIZATIONS OF AN INTEGRAL INEQUALITY
QU ˆO´ C ANH NG ˆO AND FENG QI
Abstract. In this short paper, an integral inequality posed in the 11th Inter- national Mathematical Competition for University Students, Skopje, Macedo- nia, 25–26 July 2004 is generalized.
1. introduction
The Problem 2 of the 11th International Mathematical Competition for Univer- sity Students, Skopje, Macedonia, 25–26 July 2004 (see [1]) reads as follows.
Proposition 1([1]). Letf, g: [a, b]→[0,∞)be two continuous and non-decreasing functions such that
Z x
a
pf(t) dt≤ Z x
a
pg(t) dt (1)
forx∈[a, b]and
Z b
a
pf(t) dt= Z b
a
pg(t) dt. (2)
Then
Z b
a
p1 +f(t) dt≥ Z b
a
p1 +g(t) dt. (3)
It is clear that, considering (2), inequality (1) can be rewritten as Z b
x
pf(t) dt≥ Z b
x
pg(t) dt. (4)
If replacing f(x) by p
f(x) and g(x) by p
g(x) , then Proposition 1 can be simplified into the following Proposition 2.
Proposition 2. Let f, g : [a, b] → [0,∞) be two continuous and non-decreasing functions such that
Z b
x
f(t) dt≥ Z b
x
g(t) dt (5)
forx∈[a, b]and
Z b
a
f(t) dt= Z b
a
g(t) dt. (6)
2000Mathematics Subject Classification. 26D15.
Key words and phrases. Integral inequality, generalization, the second mean value theorem for integrals.
1
2 Q. A. NG ˆO AND F. QI
Then
Z b
a
p1 +f2(t) dt≥ Z b
a
p1 +g2(t) dt. (7)
If denoting
F(x) = Z x
a
f(t) dt and G(x) = Z x
a
g(t) dt, (8)
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 (7) is valid apparently, because the length of the curve y =F(x) is not less than that of the curvey=G(x). This explains the geometric meaning of Proposition 2 and gives a solution to Proposition 1.
The main aim of this paper is to generalize Proposition 1 and Proposition 2 above.
Our main results are included in a couple of theorems below.
Theorem 1. Let f : [a, b]→[0,∞)be a continuous function andg: [a, b]→[0,∞) a continuous and non-decreasing function satisfying (5)and (6). Then
Z b
a
h(f(t)) dt≥ Z b
a
h(g(t)) dt (9)
validates for every convex functionhon[0,∞).
Theorem 2. Let f : [a, b]→[0,∞)be a continuous function andg: [a, b]→[0,∞) a continuous and non-increasing function satisfying (6) and the reverse of (5).
Then inequality (9)holds true for every convex functionhon [0,∞).
Remark 1. It is easy to see that the function √
1 +x2 is convex on [a, b]. Hence, inequality (7) can be deduced easily from (9).
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 availble.
Lemma 1 ([2, p. 35]). Letf(x)be bounded and monotonic andg(x)integrable on [a, b]. Then there exists some ξ∈[a, b]such that
Z b
a
f(x)g(x) dx=f(a) Z ξ
a
g(x) dx+f(b) Z b
ξ
g(x) dx. (10)
Proof of Theorem 1. Denote φ(x) =−
Z b
x
f(t) dt and ϕ(x) =− Z b
x
g(t) dt. (11)
Sincehis convex, then
h(t)≥h(s) + (t−s)h0(s) fora≤s, t≤b, therefore,
h(φ0(t))≥h(ϕ0(t)) + [φ0(t)−ϕ0(t)]h0(ϕ0(t)) (12) which gives
Z b
a
h(φ0(t)) dt≥ Z b
a
h(ϕ0(t)) dt+ Z b
a
[φ0(t)−ϕ0(t)]h0(ϕ0(t)) dt. (13)
Now it is sufficient to prove that Z b
a
[φ0(t)−ϕ0(t)]h0(ϕ0(t)) dt≥0. (14) Sinceh(t) is convex, the functionh0(t) is non-decreasing; Sinceg(t) is non-decreasing, the functionϕ0(t) is also non-decreasing. Thus, the composite functionh0(ϕ0(t)) is non-decreasing with respect tot. Using Lemma 1 yields
Z b
a
[φ0(t)−ϕ0(t)]h0(ϕ0(t)) dt
=h0(ϕ0(a)) Z ξ
a
[φ0(t)−ϕ0(t)] dt+h0(ϕ0(b)) Z b
ξ
[φ0(t)−ϕ0(t)] dt
=h0(g(a))[φ(ξ)−φ(a)−ϕ(ξ) +ϕ(a)] +h0(g(b))[φ(b)−φ(ξ)−ϕ(b) +ϕ(ξ)]
= [φ(ξ)−ϕ(ξ)][h0(g(a))−h0(g(b))]
≥0,
whereξ∈[a, b], sinceφ(ξ)≤ϕ(ξ) andh0(g(a))≤h0(g(b)). The proof of Theorem 1
is complete.
Proof of Theorem 2. Denote ψ(x) =
Z x
a
f(t) dt and θ(x) = Z x
a
g(t) dt (15)
fora≤x≤b. Along with the proof of Theorem 1, it is obtained that, in order to show Theorem 2, it suffices to prove
Z b
a
[ψ0(t)−θ0(t)][−h0(θ0(t))] dt≤0. (16) Since h is convex, then −h0 is non-increasing; Since g is non-increasing, then θ0 is also non-increasing. Consequently, the composite function −h0(θ0(t)) is non- increasing with respect tot. Utilizing Lemma 1 leads to
Z b
a
[ψ0(t)−θ0(t)]h0(θ0(t)) dt
=h0(θ0(a)) Z ξ
a
[ψ0(t)−θ0(t)] dt+h0(θ0(b)) Z b
ξ
[ψ0(t)−θ0(t)] dt
= [ψ(ξ)−θ(ξ)][h0(g(a))−h0(g(b))]
≥0,
whereξ∈[a, b], sinceψ(ξ)≥θ(ξ) andh0(g(a))≥h0(g(b). The proof of Theorem 2
is complete.
References
[1] The 11th International Mathematical Competition for University Students, Skopje, Mace- donia, 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 and B. C. Drachman,Inequalities With Applications to Engineering, Springer, 1998.
4 Q. A. NG ˆO AND F. QI
(Q. A. Ngˆo)Department of Mathematics, Mechanics and Informatics, College of Sci- ence, Viˆe. t Nam National University, H`a Nˆo. i, Viˆe. t Nam
E-mail address:bookworm [email protected]
(F. Qi)Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China
E-mail address:[email protected], [email protected], [email protected] URL:http://rgmia.vu.edu.au/qi.html
