Mean value theorem for the second derivative, when the first derivative is zero at endpoints [duplicate]

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.