Proving that $(\sup_{x\in R}|f'(x)|)^2\leq 4\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$. [duplicate]

I was google-ing and came across this question. Till now I don't have any solution.

Let $f$ be a double differentiable function on $(1,\infty)$. Let $M_0=\sup_{x\in R}|f(x)|$, $M_1=\sup_{x\in R}|f'(x)|$ and $M_2=\sup_{x\in R}|f''(x)|$. Prove that $M_1^2\leq 4M_0M_2$.

The only ideas that are striking me are Mean Value Theorem and Taylor's Theorem. How can this be solved?

Solution 1:

Let's note a few things. If any of $M_0, M_1, M_2$ are zero, then this is trivial. In particular, $M_0 = 0$ means $f \equiv 0$. $M_1 = 0$ is similarly trivial. If $M_2 = 0$, then $f$ is a linear function, so either $f \equiv c$, in which case the inequality holds, or $f = ax + b$ and $M_0$ is infinite, contradicting that each $M_0, M_1, M_2$ are finite. (If they don't need to be finite, then the inequality isn't true - by taking linear functions).

So we have $M_0, M_1, M_2 > 0$. Taylor's Theorem gives us $\xi \in (x, x + \theta)$ such that $$ f(x + \theta) = f(x) + f'(x)\theta + \frac{1}{2}f''(\xi)\theta^2.$$ Rearranging, $$ f'(x) = \frac{1}{\theta} (f(x+\theta) - f(x)) - \frac{\theta}{2} f''(\xi),$$ so that $$ \lvert f'(x) \rvert \leq \frac{1}{\theta} \left( \lvert f(x + \theta) \rvert + \lvert f(x) \rvert\right) + \frac{\theta}{2} \lvert f''(\xi) \rvert \leq \frac{2}{\theta}M_0 + \frac{\theta}{2} M_2.$$ Take $\theta = \dfrac{2\sqrt M_0}{\sqrt M_2}$ to see that $$ \lvert f'(x)\rvert \leq 2\sqrt{M_0M_2}.$$ Taking sups and squaring gives $$ M_1^2 \leq 4M_0M_2,$$ as we wanted to show. $\diamondsuit$