Prove that $|f''(\xi)|\geqslant\frac{4|f(a)-f(b)|}{(b-a)^2}$ Let ${\rm f}:\left[a, b\right]\to\mathbb{R}$ be twice differentiable, and suppose 
$$\lim_{x\to a^{+}}
\frac{{\rm f}\left(x\right) - {\rm f}\left(a\right)}{x - a}
=
\lim_{x\to b^{-}}\frac{{\rm f}\left(x\right) - {\rm f}\left(b\right)}{x - b}
=0
$$
Show that there exists $\xi \in \left(a, b\right)$ such that
$\displaystyle{%
\left\vert\vphantom{\Large A}\,{\rm f}''\left(\xi\right)\right\vert
\geq
\frac{4\left\vert\vphantom{\Large A}%
\,{\rm f}\left(a\right) - {\rm f}\left(b\right)\right\vert}
{\left(b - a\right)^{2}}}$.
I don't know how to start. Any hints ?.
 A: let $g(x) = f'(x)$, then with $g(a) = g(b)= 0$ we are going to prove 
$$|\int_{a}^{b}g(x)dx| \leq\left(\sup_{\xi \in (a,b)}|g'(\xi)|\right)\frac{(b-a)^2}{4}$$
\begin{align}
|\int_{a}^{\frac{a+b}{2}}g(x)dx| &= |\int_{a}^{\frac{a+b}{2}}g(x)-g(a)dx| \\
&\leq \left(\sup_{\xi \in (a,b)}|g'(\xi)|\right)\int_{a}^{\frac{a+b}{2}}(x-a)dx \\
&= \left(\sup_{\xi \in (a,b)}|g'(\xi)|\right) \frac{(b-a)^2}{8}
\end{align}
Similarly we could also get $|\int_{\frac{a+b}{2}}^{b}g(x)dx|  \leq\left(\sup_{\xi \in (a,b)}|g'(\xi)|\right) \frac{(b-a)^2}{8}$, then the result follows by adding the two inequalities
A: An alternative, more enlightening proof that follows what has been done here https://math.stackexchange.com/a/835185/66096
For variational purposes, we assess the integral $\displaystyle \int_a^bf''(t)(t+\beta)$dt where $\beta$ is arbitrary.
By integration by parts, and since $f'(a)=f'(b)=0$, this rewrites as $\displaystyle \int_a^bf''(t)(t+\beta)dt =f(a)-f(b)$
Hence $\displaystyle \max|f''|\int_a^b|t+\beta|dt\geq \left|\int_a^bf''(t)(t+\beta)dt \right| =|f(b)-f(a)| $
And $\displaystyle \max|f''|\geq \frac{|f(b)-f(a)|}{\int_a^b|t+\beta|dt}$
Choosing $\displaystyle \beta=-\frac{a+b}{2}$ yields the result.
