You have $f \in C^\infty([0,1])$ with $f > 0$. Then $\sqrt{f}$ is easily seen to be differentiable . Prove that there exists a constant $C$ independent of $f$ such that:
\sup_{x\in[0,1]}\left\lvert \left(\sqrt f\right)'(x)\right\rvert \leq C \left(1 + \sup_{x\in[0,1]}\lvert f(x)\rvert + \sup_{x\in[0,1]}\lvert f'(x)\rvert + \sup_{x\in[0,1]} \lvert f''(x)\rvert\right).
I wonder if someone can give a nice proof of this.