Prove $(b-a)\cdot f(\frac{a+b}{2})\le \int_{a}^{b}f(x)dx$ Let $f$ be continuously differentiable on $[a,b]$. If $f$ is concave up, prove that
$$(b-a)\cdot f\left(\frac{a+b}{2}\right)\le \int_{a}^{b}f(x)dx.$$
I know that (and have proved) $$(b-a)\cdot f\left(\frac{a+b}{2}\right)= \int_{a}^{b}f(x)dx$$ for any linear function on $[a,b]$. Also, the graph of $f$ lies above the tangent line at $(\frac{a+b}{2},f(\frac{a+b}{2}))$.

Let $h(x)=f(x)-m x$ where $m=f'(\frac{b+a}2)$.  By the Mean Value Theorem for integrals (also related to average value of a function), there is $c\in(a,b)$ such that $h(c)(b-a)=\int_{a}^{b}h(x)dx$. Note that $h$ is concave up and $h'(\frac{b+a}2)=0$ hence $h$ has a global (over $[a,b]$) minimum at $\frac{b+a}2$. Hence $h(c)\ge h(\frac{b+a}2)$. In other words $\frac1{b-a}\int_{a}^{b}h(x)dx\ge h(\frac{b+a}2)$. Thus $\frac{1}{b-a}\int_{a}^{b}h(x)dx = \frac1{b-a}(\int_{a}^{b}f(x)dx - \frac{m}{2} (b^2-a^2))= \frac1{b-a}\int_{a}^{b}f(x)dx - {m}\frac{b+a}{2}\ge$ $h(\frac{b+a}2)=f(\frac{b+a}2)-{m}\frac{b+a}{2}$ 
from which we obtain $\frac1{b-a}\int_{a}^{b}f(x)dx\ge f(\frac{b+a}2)$. 
