Prove that $|f'(x)|\leq\sqrt{2MM''}$ Let $f:\mathbb{R}\to\mathbb{R}$ be twice differentiable with
$$|f(x)|\leq M, |f''(x)|\leq M'',\forall x\in\mathbb{R}$$
Prove that $|f'(x)|\leq\sqrt{2MM''},\forall x\in\mathbb{R}$
I am thinking about using Taylor's theorem:
For any $x\in\mathbb{R}$ and $a>0$, by Taylor's theorem $\exists \xi\in(x,x+a)$ s.t.
$$f(x+a)=f(x)+f'(x)a+\frac{f''(\xi)}{2}a^2$$
Thus
$$|f'(x)|\leq\frac{|f(x)|+|f(x+a)|}{a}+\frac{|f''(\xi)|}{2}a$$
However with this approach the best bound we can get is
$$|f'(x)|\leq 2\sqrt{MM''}$$
Thus I feel that there is probably a completely different trick. 

A variant which is basically equivalent but with a slightly more geometric touch calculates a minimum bound for M in terms of $f'(x)$. 
Assume without loss of generality that $f(a)\ge 0$ and $f'(a) > 0$. 
For $x > a$ we find from integrating the lower bound of the second derivative that $$f'(x) \geq f'(a) - M''(x-a)$$
Integrating again from a to x, we get 
$$f(x) \geq f'(a)(x-a) - M'' \frac{(x-a)^2}{2} $$
This second degree polynomial has a maximum when $x-a = \frac{f'(a)}{M''} $ and the corresponding value is $\frac{f'(a)^2}{2M''}$, giving us 
$$M \geq f(x) \geq \frac{f'(a)^2}{2M''} $$
 This is equivalent to the inequality that should be proved.
