Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| $ [closed]

Let $f\left( x\right)$ be a $C^{2}$ function on $\mathbb{R}$. Show that $$\sup \left| f'\left( x\right) \right| ^{2}\leqslant4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| .$$

Let $\sup|f^{(n)}(x)|=M_n$. Since $f$ is $C^2$-continuous, we can use Taylor's theorem to write:

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}2f''(t)$$

for some $x<t<x+h$. Then we can solve for $f'(x)$:

$$f'(x)=\frac1h(f(x+h)-f(x))-\frac h2f''(t)$$ $$|f'(x)|\leq\frac1h(|f(x+h)|+|f(x)|)+\frac h2|f''(t)|\leq\frac2hM_0+\frac h2M_2.$$

Since we can apply this for any $h$, choose $h=2\sqrt{M_0/M_2}$, so that

$$|f'(x)|\leq\sqrt{M_2/M_0}\cdot M_0+\sqrt{M_0/M_2}\cdot M_2=2\sqrt{M_0M_2}.$$

Since this inequality is true for any $f'(x)$, it is true for the supremum, i.e.

$$M_1\leq2\sqrt{M_0M_2}\Rightarrow\sup|f'(x)|^2\leq4\sup|f(x)|\sup|f''(x)|.$$