Show that $f\left(\frac{x_1+x_2}{2}\right)\leq\frac{f(x_1)+f(x_2)}{2}$ Let $f$ be twice differentiable in $(a,b)$ and $f''>0$ in the same interval. If $a<x_1<x_2<b,$ show that $$f\left(\frac{x_1+x_2}{2}\right)\leq\frac{f(x_1)+f(x_2)}{2}.$$
I'm not sure how to start. I know at least that $f'$ is strictly increasing in $(a,b).$ Should I check separately for the cases $x_1\neq x_2$ and $x=x_2$? I'm supposed to use the mean value theorem, but I don't know how. Any insight I need to make in order to start? 

What you want here is the convexity of $f$. Since you know that $f'$ is monotonically increasing, I will outline a proof for you of how to show $f$ is convex.
Proof outline. Let $a< x < y < b$ and consider the slope $m$ of the line $L$ joining the points $(x,fx)$ and
$(y,fy)$. By the mean value theorem, there exists a point $c\in (x,y)$ such that
$f'(c) = m$. Suppose that there is a point $\theta\in(x,y)$ such that $f(\theta)>L(\theta)$. Consider separately the two cases where $\theta\in(x,c)$ and $\theta\in(c,y)$ and
derive contradictions. (Draw pictures.)
