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. 

Hint: You have
\begin{equation*}
 f'(x) = \frac{f(x+a)-f(x)}{a} + \frac{a}2 f''(\xi).
\end{equation*}
Let us assume that $f'(x) \ge 0$. Then,
\begin{align*}
f'(x)
& \le \frac{|f(x+a)|}{|a|}-\frac{f(x)}{a} + \frac{|a|}2 |f''(\xi)|.
\\
& \le \frac{M}{|a|}-\frac{f(x)}{a} + \frac{|a|}2 M''.
\end{align*}
Now the trick is to choose the sign of $a$, such that $-\frac{f(x)}{a} \le 0$.
Edit: A function which attains this bound can be constructed as follows: Set
$$f''(x) = \begin{cases}-1 & x \in [0,\sqrt{2})\\ -1 + (x - \sqrt{2}) & x \in [\sqrt{2}, \frac32\,\sqrt{2}+1) \\ \frac12 \sqrt{2} + (\frac32 \, \sqrt{2}+1-x) & x \in [\frac32 \, \sqrt{2} +1 , 2\sqrt{2}+1) \\
0 & x \ge 2\sqrt2+1\end{cases}$$
Together with $f(x) = 0$ and $f'(x)=\sqrt2$ we find $f(x)$ for $x \ge 0$ and we set $f(-x ) = -f(x)$. Then, $M = M'' = 1$, but $f'(0) = \sqrt2$.
