Proving that an inequality is true, from assuming that second derivatives exist, and first derivatives are zero on the boundary, EDIT 2: I just posted my revised proof, where I used two Taylor expansions, and subtracting both equations to get something that's pretty close to what I want.  What do you think?  Please see below.  Thanks,
EDIT:  I think that, with the help of Joey Zou and Claudeh5's attempts at a solution (please see below), I am pretty close to an answer.  At present there are some things of concern:
a) Joey Zou's more technical proof seems to rely on $f''$ being continuous or at least Riemann-integrable.  Unfortunately, I don't think we can assume that.
b) Claudeh5's cute one-line proof almost does the job.  It uses Taylor expansion, the Lagrange Remainder, and centering the expansion about  $x=a$, and evaluating the series at $x=b$.  However, the estimate is not quite good enough.
Any hints or comments are welcome.  Thanks,
The problem statement is:
Assume that $f(x)$ has second derivatives on [a,b], and $f′(a)=f′(b)=0$. 
Prove that there exists a point $c∈[a,b]$ such that $f′′(c)≥\frac{4}{(a−b)^2}|f(b)−f(a)|$.
Using Claudeh5's approach, here is my revised proof, which is perhaps closer to the desired upper bound:
By Taylor expansion, and the Lagrange remainder, we have that 
$$f(b) = f(a) + \frac{f''(\psi_1)}{2!} (b-a)^2$$
Now again, but centering the expansion about $x=b$ gives us 
$$f(a) = f(b) + \frac{f''(\psi_2)}{2!} (a-b)^2$$
Note that the first derivatives vanish at $x=a$ and $x=b$, by assumption.
Now subtracting the two Taylor series gives
$$2[f(b)-f(a)]= \frac {[f''(\psi_1) - f''(\psi_2)]}{2!}(a-b)^2$$
$$\implies \frac  {4[f(b)-f(a)]}{(a-b)^2}= [f''(\psi_1) - f''(\psi_2)]$$
$$\implies \frac  {4[f(b)-f(a)]}{(a-b)^2} \le max \{f''(\psi_1), f''(\psi_2)\}$$
$$=:f''(c)$$
I am not confident about the last two lines of my proof.
Am I on the right track?  I feel a bit closer now to achieving the upper bound ...
Thanks,
 A: hint2 : with Taylor-Lagrange formula : $f(b)=f(a) +(b-a) f'(a)+ \frac{(b-a)^2}2 f''(d)$ ($d$ \in ]a,b[$)
so here, $f(b) - f(a) = \frac{(b-a)^2}2 f''(d)$ and if $M=\max_{d \in[a,b]} |f''(d)|$, $\exists c$ with $|f''(c)| = M$.
so, $|f(b)-f(a)| = \frac{(b-a)^2}2 |f''(d)| \le \frac{(b-a)^2}2 |f''(c)|$ 
but we have only 2, not 4...
A: Let $M = \sup\limits_{x\in[a,b]}{|f''(x)|}$. Since $f'(a) = 0$, we have, for $a\le x\le b$,
$$ |f'(x)| = \left|f'(a) + \int\limits_{a}^{x}{f''(x)\text{ d}x}\right|\le\int\limits_{a}^{x}{|f''(x)|\text{ d}x}\le M(x-a). $$
Since $f'(b) = 0$, we similarly have
$$ |f'(x)| = \left|f'(b) + \int\limits_{x}^{b}{f''(x)\text{ d}x}\right|\le\int\limits_{x}^{b}{|f''(x)|\text{ d}x}\le M(b-x). $$
It follows that
\begin{align*} |f(b)-f(a)| &= \left|\int\limits_{a}^{b}{f'(x)\text{ d}x}\right|\\
&\le\int\limits_{a}^{\frac{a+b}{2}}{|f'(x)|\text{ d}x} + \int\limits_{\frac{a+b}{2}}^{b}{|f'(x)|\text{ d}x}\\
&\le\int\limits_{a}^{\frac{a+b}{2}}{M(x-a)\text{ d}x} + \int\limits_{\frac{a+b}{2}}^{b}{M(b-x)\text{ d}x} \\
&= M\frac{(b-a)^2}{4}.
\end{align*}
Furthermore, I claim that equality cannot be achieved. Note that $f'$ is continuous (as $f''$ exists), and since $|f'(x)|\le M(x-a)$, the only way for the equality
$$\int\limits_{a}^{\frac{a+b}{2}}{|f'(x)|\text{ d}x} = \int\limits_{a}^{\frac{a+b}{2}}{M(x-a)\text{ d}x}$$
to hold is if $|f'(x)| = M(x-a)$ for all $x\in\left(a,\frac{a+b}{2}\right)$. Similarly, for the other equality to hold, we need $|f'(x)| = M(b-x)$ for all $x\in\left(\frac{a+b}{2},b\right)$. But it is impossible for $f'$ to be differentiable at $\frac{a+b}{2}$ while satisfying these two equalities. Hence, the inequality is strict, so
$$|f(b)-f(a)| < M\frac{(b-a)^2}{4}\implies M > \frac{4}{(b-a)^2}|f(b)-f(a)|. $$
Since $M$ is the supremum over all values of $|f''(x)|$ on $[a,b]$, it follows that there exists $c\in[a,b]$ such that $|f''(c)|\ge\frac{4}{(b-a)^2}|f(b)-f(a)|$.

Edit: as pointed out by Lebron James, it is not necessarily true that $f''$ is Riemann integrable, or even bounded. Now


*

*If $f''$ is not bounded, then we automatically get our result.

*Otherwise, $M=\sup\limits_{x\in[a,b]}{|f''(x)|}$ exists and is finite. We can still show that $|f'(x)|\le M(x-a)$ and $|f'(x)|\le M(b-x)$ using the mean value theorem. The rest of thep proof should still work.
