proof of Hermite-Hadamard integral inequality

First of all, let’s recall that a convex function on a openinterval $(a,b)$ is continuous on $(a,b)$ and admits left and rightderivative $f^{+}(x)$ and $f^{-}(x)$ for any $x\in (a,b)$. For this reason,it’s always possible to construct at least one supporting line (http://planetmath.org/ConvexFunctionsLieAboveTheirSupportingLines) for $f\left(x\right)$ at any $x_0\in (a,b)$ : if $f\left(x_0\right)$ isdifferentiable in $x_0$, one has $r(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)$; if not, it’s obvious that all $r(x)=f\left(x_0\right)+c\left(x-x_0\right)$ are supporting lines forany $c\in [f^{-}(x_0),f^{+}(x_0)]$.
Let now $r(x)=f\left(\frac{a+b}{2}\right)+c\left(x-\frac{a+b}{2}\right)$ be a supporting line of $f(x)$ in $x=\frac{a+b}{2}\in (a,b)$.Then, $r\left(x\right)\le f\left(x\right)$. On the other side, byconvexity definition, having defined $s\left(x\right)=f\left(a\right)+\frac{f(b)-f(a)}{b-a}(x-a)$ the line connecting the points $(a,f(a))$ and $(b,f(b))$ , one has $f(x)\le s(x)$. Shortly,

$$r\left(x\right)\le f\left(x\right)\le s(x)$$

Integrating both inequalities between $a$ and $b$

$${\int }_a^{b}r\left(x\right)𝑑x\le {\int }_a^{b}f\left(x\right)𝑑x\le {\int }_a^{b}s(x)𝑑x$$

${\displaystyle {\int }_a^b}r\left(x\right)𝑑x$

$=$${\displaystyle {\int }_a^b}\left[f\left({\displaystyle \frac{a+b}{2}}\right)+c\left(x-{\displaystyle \frac{a+b}{2}}\right)\right]𝑑x$

$=$$f\left({\displaystyle \frac{a+b}{2}}\right)(b-a)+c{\displaystyle {\int }_a^b}\left(x-{\displaystyle \frac{a+b}{2}}\right)𝑑x$

$=$$f\left({\displaystyle \frac{a+b}{2}}\right)(b-a)$

${\displaystyle {\int }_a^b}s(x)𝑑x$

$=$${\displaystyle {\int }_a^b}\left[f\left(a\right)+{\displaystyle \frac{f(b)-f(a)}{b-a}}(x-a)\right]𝑑x$

$=$$f(a)(b-a)+{\displaystyle \frac{f(b)-f(a)}{b-a}}{\displaystyle {\int }_a^b}(x-a)𝑑x$

$=$${\displaystyle \frac{f(a)+f(b)}{2}}(b-a)$

and so

$$f\left(\frac{a+b}{2}\right)(b-a)\le {\int }_a^{b}f\left(x\right)𝑑x\le \frac{f(a)+f(b)}{2}(b-a)$$

which is the thesis.