TITLE: If $f$ is strictly increasing and strictly convex, is $f'(x)^2 > f(x) f''(x)$ for all $x$?
QUESTION [5 upvotes]: Question
Suppose $f: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ with $f(0) = 0$, $f'(x) > 0$ for all x, and $f''(x) > 0$ for all $x$.
Is $f'(x)^2 > f(x) f''(x)$ for all $x$?
My Attempts
Attempt 0: This is true for the special case in which $f(x) = x^\alpha$ for $\alpha \geq 2$.  $\alpha^2 x^{2 \alpha - 2} > \alpha (\alpha - 1) x^{2 \alpha - 2}$, which reduces to $1 > 0$.
Attempt 1: $f'(0)^2 > f(0) f''(0)$ because $f'(0) > 0$ and $f(0) = 0$.  If the derivative of the left hand side is weakly greater than the derivative of the right hand side, then the proposition is true.  The LHS derivative is weakly greater if $f'(x) f''(x) > f(x) f'''(x)$, but we don't know anything about $f'''(x)$.
Attempt 2: Since we know this is true for $x = 0$, we can focus on the case where $x > 0$.  For $x > 0$, this is proposition is equivalent to $\frac{f'(x)}{f(x)} > \frac{f''(x)}{f'(x)}$.  This is equivalent to $\frac{d}{dx} \log f(x) > \frac{d}{dx} \log f'(x)$.  Using the second fundamental theorem of calculus and $f(0) = 0$, $f(x) = f(x) - f(0) = \int_{0}^{x} f'(y) dy$.  Substituting, we need $\frac{d}{dx} \log \int_{0}^{x} f'(y) dy > \frac{d}{dx} \log f'(x)$.  This seems like it should be useful, but I wasn't able to figure out how to proceed from here.

REPLY [2 votes]: Even the non-strict inequality does not hold.
Take $f(x) = e^{x^2}-1$. Then $f$ is increasing and strictly convex and $f(0) = 0$. We have $f'(x) = 2x e^{x^2}$ and $f''(x) = (2 + 4x^2)e^{x^2}$. Thus
$f(x) f''(x) = (2+4x^2) e^{2x^2} - (2+4x^2)e^{x^2}$ and $f'(x)^2 = 4x^2 e^{2x^2}$. So, we have
$$\forall x \geqslant 0,\ f'^2(x)-f(x)f''(x) = (2+4x^2)e^{x^2}-2e^{2x^2}.$$
The right-hand side is clearly negative for large $x$.