Mean value theorem for the second derivative, when the first derivative is zero at endpoints Suppose $f:[a,b]\to \mathbb R$ has derivative up to order $2$, and 
$f'(a)=f'(b)=0$.
Prove there is  a point $c$ at $(a,b)$ such that 
$$
|f''(c)|\geq 4\frac{|f(b)-f(a)|}{(b-a)^2}.
$$

If it was factor 2, not 4, then I could use a Taylor expansion with Lagrange residue.

Let $z = \frac{a + b}2$.
Taylor's expansion centered at $a$ is
$$
f(z) = f(a) + f'(a)(z - a) + \frac 12f''(\xi)(z - a)^2
$$
for some $\xi \in [a, z]$.
Since $f'(a) = 0$, we get
\begin{align}
f(z) - f(a) & = \frac 12 f''(\xi)(z - a)^2\\
\therefore f''(\xi) & = 2\frac{f(z) - f(a)}{(z - a)^2}. \tag{1}
\end{align}
Similarly, expansion centered at $b$ is
\begin{align}
f(z) & = f(b) + \frac 12f''(\eta)(z - b)^2\\
\therefore
f''(\eta) & = 2 \frac{f(z) - f(b)}{(z - b)^2} \tag{2}
\end{align}
for some $\eta \in [z, b]$.
Subtract (2) from (1), and divide by $2$:
\begin{align}
\frac{f''(\xi) - f''(\eta)}{2} = 4\frac{f(b) - f(a)}{(b - a)^2}.
\end{align}
Now, if $|f''(\xi)| \ge |f''(\eta)|$, it will follow that
$$
|f''(\xi)| \ge \frac{|f''(\xi)| + |f''(\eta)|}{2}
\ge \frac{|f''(\xi) - f''(\eta)|}{2} = 4\frac{|f(b) - f(a)|}{(b - a)^2}.
$$
Otherwise, $|f''(\xi)| < |f''(\eta)|$, and we get
$$
|f''(\eta)| \ge 4\frac{|f(b) - f(a)|}{(b - a)^2}
$$
by a similar reasoning.
