If \(\displaystyle{x\in\left[a,b\right]}\), then \(\displaystyle{a+bx\in\left[a,b\right]}\).
Since the function \(\displaystyle{f}\) is convex, we get :
\(\displaystyle{f(x)+f(a+bx)\geq 2\,f\,\left(\dfrac{x+a+bx}{2}\right)=2\,f\,\left(\dfrac{a+b}{2}\right)\,,\forall\,x\in\left[a,b\right]}\) .
The function \(\displaystyle{f}\) is continuous, so by integration we have that :
\(\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x+\int_{a}^{b}f(a+bx)\,\mathrm{d}x\geq 2\,(ba)\,f\,\left(\dfrac{a+b}{2}\right)}\)
or :
\(\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x\int_{b}^{a}f(a+bx)\,\mathrm{d}(a+bx)\geq 2\,(ba)\,f\,\left(\dfrac{a+b}{2}\right)}\)
or :
\(\displaystyle{2\,\int_{a}^{b}f(x)\,\mathrm{d}x\geq 2\,(ba)\,f\,\left(\dfrac{a+b}{2}\right)}\)
which means that :
\(\displaystyle{f\,\left(\dfrac{a+b}{2}\right)\leq \dfrac{1}{ba}\,\int_{a}^{b}f(x)\,\mathrm{d}x}\) .
On the other hand, from the first mean value theorem,
\(\displaystyle{\dfrac{1}{ba}\,\int_{a}^{b}f(x)\,\mathrm{d}x=f(\xi)}\) for some \(\displaystyle{\xi\in\left(a,b\right)}\) .
Please, give a hint for the righthand inequality.
Geometrical comment
We have that :
\(\displaystyle{\int_{a}^{b}f\,\left(\dfrac{a+b}{2}\right)\,\mathrm{d}x\leq \int_{a}^{b}f(x)\,\mathrm{d}x\leq \int_{a}^{b}\dfrac{f(a)+f(b)}{2}\,\mathrm{d}x}\)
which means that the area of the parallelogram defined by
\(\displaystyle{A(a,0),B(b,0),C\left(a,f\left(\dfrac{a+b}{2}\right)\right),D\left(b,f\left(\dfrac{a+b}{2}\right)\right)}\)
is less or equal to the area of the region defined by the graph of the function \(\displaystyle{f}\) and
the lines \(\displaystyle{x=a\,,x=b}\) and this area is less or equal to the area of the table
defined by \(\displaystyle{E(a,0),F(a,f(a)),G(b,0),H(b,f(b))}\) .
