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. 

Edit: Oops, someone already said this in a comment. No harm in leaving this up - no votes, please.
Old trick: Consider
$$f(x+a)=f(x)+f'(x)a+\frac{f''(\xi_1)}{2}a^2$$
and 
$$f(x-a)=f(x)-f'(x)a+\frac{f''(\xi_2)}{2}a^2.$$
Subtract:
$$|f'(x)|\le\frac Ma+\frac a2M''.$$
Let $a=\sqrt{2M/M''}$.
