Q:

Proving a property of a twice differentiable function with the mean value theorem

Let $f: \mathbb{R} \to \mathbb{R}$ is a twice differentiable function, and $a,b$ are two real numbers such that $a < b$ and $f(a) = f(b) = 0$. Prove that if $f''(x) > 0$ for all $x \in (a,b)$ then $f(x) < 0$ for all $x \in (a,b)$.
My progress so far:
Since $f''(x) > 0$, we have $f'(x)$ is an increasing function. Hence $f'(a)\le f'(x_0) \le f'(b)$ for all $x_0 \in (a,b)$. But $f'(a) = f'(b)$, so $f'(x)$ is a constant for $x \in (a,b)$.
On the other hand, by mean value theorem $\exists c \in (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a} = 0$. 
Hence we can conclude that $f'(x) = 0$ for all $x \in (a,b)$, or $f(x) = C$ a constant in $(a, b)$.
Problem:
I stuck after that point and don't know what to do next. Any suggestion?

A:

Since $f'(x)$ is strictly increasing, by the mean value theorem,  there exist $a < \xi_1 < x<\xi_2 < b$ such that
$$\tag{*}\frac{f(x)- f(a)}{x-a} = f'(\xi_1) < f'(\xi_2) = \frac{f(b) - f(x)}{b-x} $$
Rearranging the inequality (*) we have
$$f(x) \frac{b-x}{x-a} - f(a) \frac{b-x}{x-a} < f(b) - f(x) \\ \implies f(x) \frac{b-a}{x-a} < f(a) \frac{b-x}{x-a} + f(b)\\ \implies f(x) < f(a)\frac{b-x}{b-a} + f(b) \frac{x-a}{b-a} = 0$$

