Prove Jensen’s inequality: $F(\frac{1}{μ(X)}\int f \,dμ) ≤ \frac{1}{μ(X)} \int F(f)\,dμ.$ Let $(X,A,μ)$ be a finite measure space, and let $F : \mathbb{R} → \mathbb{R}$ be a $C^2$ function with second derivative $F'' > 0$. Let $f \in L_1(\mu)$ be real-valued. Prove Jensen’s inequality:
$$F\left(\frac{1}{μ(X)}\int f \,d\mu\right) \le \frac{1}{\mu(X)} \int F(f)\,d\mu.$$
I am sure this isn't too difficult, but I am not so sure how to begin, and I can't find the proof in this context online.  I would appreciate some help.  It a practice qual prob.  Thanks.

$F^{\prime\prime}>0$ means the function is convex (if $a<b<c$ are such that $\frac{F(a)-F(b)}{a-b}>\frac{F(c)-F(b)}{c-b}$ then apply mean value theorem twice to find that there exist a point where $F^{\prime\prime}<0$). The rest of this answer is just going to be standard proof of Jensen's inequality.
Let $E=\frac{1}{\mu(X)}\int fd\mu$. Since $F$ is convex, there exist a slope $m$ such that $F(x)\geq F(E)+m(x-E)$ (set $m=\inf\limits_{x>0}\frac{F(x)-F(E)}{x-E}$, then existence of $m$ and the fact that it works are both consequence of convexity). Then $\frac{1}{\mu(X)}\int F(f)d\mu\geq\frac{1}{\mu(X)}\int F(E)+m(f-E)d\mu=\frac{1}{\mu(X)}\int F(E)d\mu+\frac{m}{\mu(X)}fd\mu-\frac{1}{\mu(X)}\int mEd\mu=F(E)+mE-mE=F(E)$
