Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality

In this paper we prove several integral inequalities and we find an upper bound of the Hermite-Hadamard inequality for a convex function on a bounded area from the plane in special cases.


Introduction
Let f be a convex function on [a, b]. Then we have the following inequality, which is called Hermite-Hadamard inequality: There are many extensions, generalizations and similar results of inequality (.). In [], Fejer established the following weighted generalization of inequality (.). In this paper, we find an upper bound for b a f (x)g(x) dx, where f is a convex function on [a, b] and g is non-negative increasing (or decreasing) on [a, b], and b a g(t) dt = . Finally, in Section  we find an upper bound for the following integral: h (x) F(x, y) dy dx.

Integral inequalities
Theorem . Let f : [a, b] → R be a differentiable convex function and g : x a g(t) dt.
We will show that H (x) ≤ . We have By the extended mean value theorem (Cauchy's theorem), we have On the other hand, by the convexity of f and decreasing of g, we obtain Since g is non-negative, Then we have By the mean value theorem (Lagrange's theorem), there exist ζ  ∈ ( a+x  , x) and ζ  ∈ (a, x) such that Hence, By the convexity of f and increasing of g, we obtain . So, Therefore, H is decreasing and H(b) ≤ H(a) = . The proof is complete.
Corollary . Let f : [a, b] → R be a convex function and g be a non-negative integrable function.
The proof is similar to the proof of theorem.

Right bidimensional Hermite-Hadamard inequality
Let us consider the bidimensional interval = Note that every convex function f : → R is co-ordinated convex, but the converse is not generally true; see []. Dragomir in [] established the following similar inequality of the Hermite-Hadamard inequality for a co-ordinated convex function on a rectangle from the plane R  .
Now, let be a convex area from the plane R  , bounded by a convex function y = h(x) and a concave function y = g(x) and x = a, x = b, such that for any x ∈ [a, b], g(x) ≥ h(x). http://www.journalofinequalitiesandapplications.com/content/2013/1/27 Also, let F be a two-variable convex function on . In [] and [], the following inequality is proved: In this paper, we want to find an upper bound for the integral For this purpose, we reach to the following integral: generally. So, in special cases, we will find an upper bound for the integral (.).

Theorem . Let be a bounded area by a convex function y = h(x) and a concave func-
Also, let F be a two-variable convex function on such that F(x, g(x)) and F(x, h(x)) are convex on [a, b]. Then one has the inequality . By the right-hand side of Hermite-Hadamard inequality (.), we have Integrating this inequality on [a, b], we obtain The proof is complete.
Proof By a similar way to the proof of Theorem ., we have The proof is complete.

Theorem . Let , g and h be defined as in Theorem .. Also, let F be a two-variable convex function on such that
∂F(x, g(x)) ∂g , g(a)) + F(a, h(a))) .

Proof
Denote Then we have Since F is convex, so it is co-ordinated convex. Hence, by the right-hand side of the Hermite-Hadamard inequality, we obtain So, On the other hand, we have

(t)h(t) dt
and using the fact Therefore, By a similar way, we obtain Thus, So, Then it follows that Thus, Now, notice that if F(x, g(x)), F(x, h(x)) were convex on [a, b], we can deduce the assertion of Theorem .. Since F is convex on , we have or By a similar way, we have Note that and So, Thus, Note that α(x) ≥ a. Therefore, H is decreasing and The proof is complete.
Remark . Notice that since g is concave and h is convex on [a, b], so g is decreasing and h is increasing on [a, b]. By the mean value theorem, we have In particular, if we have g(x) = mx + n, then g(x)-g(a) x-a In the following theorem, we find an upper bound of the Hermite-Hadamard inequality for a co-ordinated convex function. http://www.journalofinequalitiesandapplications.com/content/2013/1/27 Theorem . Let , g and h be defined as in Theorem .. Also, let F be a convex function only relative to y, that is, Proof Denote We have Since F is convex relative to y, by the right-hand side of the Hermite-Hadamard inequality, we obtain

Examples
Example . Let F(x, y) = x  + y  and be bounded by By easy calculation, we see that Example . Let F, g and h be defined as in Example .. By Theorem ., we have So, Thus, we can apply Theorem .

Competing interests
Authors declare that they have no competing interest.