Second derivative of convex decreasing functions larger than square of first derivative? Given a twice differentiable, convex, decreasing function $f\in C^2$ with $\lim_{x\to\infty} f(x)=k$, I am trying to identify some sufficient conditions for the second derivative to be larger than the square of the first derivative above some level $\bar x$, or
$$ f''(x) > f'(x)^2 \qquad x>\bar x.$$
I see that this is true of the negative exponential for $x>0$ and of the negative power function $1/x^n$ for $x^n>n/(n+1)$.
 A: You can prove the easier claim, "Every convex, decreasing function $f\in C^2(\mathbb{R})$ for which $f''(x)\le f'(x)^2$ satisfies $f(x)\to-\infty$ as $x \to \infty$".
To prove this, let $g(x):=-f'(x)$, so the statement becomes "every decreasing, positive function $g \in C^1(\mathbb{R})$ for which $g'(x)\ge -g(x)^2$ satisfies $\int_a^b g(x) \,dx \to \infty$ as $b \to \infty$. We only consider the case $g(x)\to 0$.
We argue that every function $g$ satisfying $g'(x)\ge -g(x)^2$ and $g(x_0)=y_0$ satisfies $g(x)\ge g_0(x)$ for all $x$, where $g_0$ is the function satisfying $g_0'(x)=-g(x)^2$ and $g_0(x_0)=y_0$. This follows since for every $0<y\le y_0$, we have $$g'(g^{-1}(y))\ge -y^2=g_0'(g_0^{-1}(y)).$$ That is, at any value $y$, $g$ cannot be decreasing faster than $g_0$ decreases at that $y$ value.
It is straightforward to check that if $g_0$ passes through $(x_0,y_0)$, then $g_0(x)=\frac{1}{x-x_0+y_0^{-1}}$, i.e. $f(x)=\int_a^x g_0(t)\,dt=\log(x-x_0+y_0^{-1})+c$, and thus the claim holds.
(As noted in the comments, you can't prove your statement without additional conditions. There is a counterexample given by initially taking $f(x):=e^{-x}$, then on each interval $(10^n,10^n+1)$, replace $f$ with a linear approximation. This gives $f''(x)=0$ and $f'(x)\ne 0$ at a sequence of points diverging to $\infty$).
