Problem
Assume that for all $x \in (0, +\infty)$, $f(x),f'(x),f''(x)$ exist, and $|f(x)|\leq a, |f''(x)|\leq b$. Prove that $|f'(x)|\leq 2\sqrt{ab}.$
My Proof
For any $x,h>0$，according to Taylor's formula expanding at $x$，we have$$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2,$$ where $0<x<\xi<x+h.$ Thus \begin{align*} |f'(x)h|&=|f(x+h)-f(x)-\frac{f''(\xi)}{2}h^2|\\ &\leq |f(x+h)|+|f(x)|+\frac{|f''(\xi)|}{2}h^2\\ &\leq a+a+\frac{bh^2}{2}\\ &=2a+\frac{bh^2}{2}. \end{align*} This shows the inequality $$|f'(x)|\leq \frac{2a}{h}+\frac{bh}{2}$$ holds for all $h>0$. Denote $$g(h):=\frac{2a}{h}+\frac{bh}{2},\forall h>0.$$ Then $|f'(x)|$ can not exceed the minimum value of $g(h)$. Since $$g(h)=\frac{2a}{h}+\frac{bh}{2}\geq 2\sqrt{\frac{2a}{h} \cdot \frac{bh}{2}}=2\sqrt{ab},$$ hence $$|f'(x)|\leq 2\sqrt{ab}.$$ The proof is completed.
Please rectify it if it's wrong. Hope to see other proof.
It should be a comment, but due to its worthiness, I make it a post. Since $|f'(x)| \le \dfrac{2a}{h}+ \dfrac{bh}{2}, \forall h > 0$, take $h = 2\sqrt{\dfrac{a}{b}}$, and this gives $|f'(x)| \le 2\sqrt{ab}$