> 
> **Exercise** Let $f\colon \mathbb{R} \to \mathbb{R}$ be $C^2$ and
>  nonnegative. Prove that 
> 
> 
> $$\big( f'(x)\big)^2 \le 2f(x)
>  \lVert f''\rVert\_{\infty}.$$
> 
> 
> 

I have found this innocent-looking little exercise... but I must admit I'm stuck on it. I have tried various roads, the most promising of them being integration by parts: (*Notation*: $\Delta\_hf(x)=(f(x+h)-f(x))/h$)

$$\frac{1}{h}\int\_x^{x+h} (f'(t))^2\, dt = f(x+h)\Delta\_hf'(x)+f'(x)\Delta\_hf(x)-\frac{1}{h}\int\_{x}^{x+h}f(t)f''(t)\, dt;$$

I had hoped to bound this identity from above then have $h \to 0^+$. But I got nowhere. 

Other approaches used Taylor expansions. All I got with those was a weaker estimate (that I reported [here](https://math.stackexchange.com/questions/38811/is-there-a-bounded-function-f-with-f-unbounded-and-f-bounded/44538#44538)). 

Can somebody give me a hint? I can't stop thinking at this exercise but I've got some work to do! :-)