$f(x)$ is a twice differentiable function in [a,b].
m is the minimum of f(x) in [a,b], and M is the maximum.
Assume $f(a) \neq m$ and $f(b) \neq m$
Prove there exists c in [a,b] such that:
$|f''(c)| \geq \frac{2(M-m)}{(b-a)^2}$
I tried to use the Taylor theorem on a few 'interesting' points in [a,b]. One thing in particular that seemed like a good direction was writing it in lagrange remainder form in point a:
$f(x) = f(a) + f'(a)(x-a) + \frac{1}{2} f''(c)(b-a)^2$
However, I couldn't get rid of the derivative term which got me stuck. I also tried to develop the function around the minimum point (which is somewhere in [a,b]), but this also didn't turn out algebraically fruitful.
I'd appreciate a hint or a direction for this.