TITLE: Mean Value Theorem involving second derivative
QUESTION [7 upvotes]: If $f:[a,b]\to\mathbb R$ such that $f'(x)>0,f''(x)>0$ for all $x\in[a,b]$, then $$\int_a^b e^{f(x)}\,dx\le (b-a)\frac{e^{f(b)}-e^{f(a)}}{f(b)-f(a)}$$

My progress so far:
We have $\frac{1}{b-a} \int\limits_{a}^{b} e^{f(x)} =e^{f(\epsilon)}$ for some $\epsilon \in (a,b)$.
Also, take the function $e^x$ int the RHS, and apply Lagrange MVT to get $e^c$ for some $c \in (f(a),f(b))$ SO, we are left to show that $e^{f(\epsilon)} \le e^c$ or $f(\epsilon) \le c$. I am stuck here. I don't know how to use the fact that $f''(x)>0$.

REPLY [2 votes]: Assuming $[a,b]=[0,1]$ without loss of generality, $f''\geq0$ means $f$ is convex, and thus
$$
f(x)\leq (1-x)f(0)+xf(1),
$$
or,
$$
e^{f(x)}\leq e^{f(0)}e^{x(f(1)-f(0))}.
$$
Therefore,
$$
\int_0^1e^{f(x)}\leq e^{f(0)}\int_0^1e^{x(f(1)-f(0))}=\frac{e^{f(1)}-e^{f(0)}}{f(1)-f(0)}.
$$
(Actually $f'>0$ is only used for the denominator to be nonzero)