Proving that $(\sup_{x\in R}|f'(x)|)^2\leq 4\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$. 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? 
 A: 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. $\diamondsuit$
A: Let $$\sup|f'(x)|=|f'(a)|, \quad \sup|f(x)|=|f(b)|$$
By Taylor theorem
$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(c)}{2}(x-a)^2$$
for some $c$. Hence
$$f'(a) = \frac{f(x)-f(a)}{x-a} - f''(c)\frac{x-a}{2}$$
$$|f'(a)| \le |f(b)|\frac{2}{|x-a|} + |f''(c)|\frac{|x-a|}{2}$$
Maximizing the last term, choosing a suitable value of $x$, we get
$$|f'(a)|\le 2\sqrt{|f(b)||f''(c)|}$$
