First of all, let's recall that a convex function on a open interval $(a,b)$ is continuous on $(a,b)$ and admits left and right derivative $f^{+}(x)$ and $f^{-}(x)$ for any $x\in (a,b)$ For this reason, it's always possible to construct at least one supporting line for $f\left( x\right) $ at any $x_{0}\in (a,b)$ : if $f\left( x_{0}\right) $ is differentiable 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 for any $c\in \lbrack f^{-}(x_{0}),f^{+}(x_{0})]$
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) \leq f\left( x\right) $ On the other side, by convexity 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 $f(x)\leq s(x)$ Shortly,$$ r\left( x\right) \leq f\left( x\right) \leq s(x) $$ Integrating both inequalities between $a$ and $b$ $$ \int_{a}^{b}r\left( x\right) dx\leq \int_{a}^{b}f\left( x\right) dx\leq \int_{a}^{b}s(x)dx $$ \begin{eqnarray*} &&\int_{a}^{b}r\left( x\right) dx \\ &=&\int_{a}^{b}\left[ f\left( \frac{a+b}{2}\right) +c\left( x-\frac{a+b}{2}% \right) \right] dx \\ &=&f\left( \frac{a+b}{2}\right) (b-a)+c\int_{a}^{b}\left( x-\frac{a+b}{2}% \right) dx \\ &=&f\left( \frac{a+b}{2}\right) (b-a) \\ && \\ &&\int_{a}^{b}s(x)dx \\ &=&\int_{a}^{b}\left[ f\left( a\right) +\frac{f(b)-f(a)}{b-a}(x-a)\right] dx \\ &=&f(a)(b-a)+\frac{f(b)-f(a)}{b-a}\int_{a}^{b}(x-a)dx \\ &=&\frac{f(a)+f(b)}{2}(b-a) \end{eqnarray*}and so$$ f\left( \frac{a+b}{2}\right) (b-a)\leq \int_{a}^{b}f\left( x\right) dx\leq \frac{f(a)+f(b)}{2}(b-a) $$ which is the thesis.