In short, I have a function $p(x)$ which is defined for $x\geq0$. The function is convex, decreasing and assymptotically approaches 0 for $x\rightarrow\infty$ . I need to know if $p''p\geq (p')^2$ .
As for the function, you may think that $p(x)$ is a type of investment function ($x$ is investment amount) and $p$ is the probability of some threat happening (the higher is the investment the lower is the probability): $p'(x)<0$. And the effectiveness of investments falls with investment, so $p''(x)>0$. If required, it is possible to assume, that the function is smooth, and $p^{(2n)}>0$ and $p^{(2n+1)}<0$ for any $n=0,1,\dots$. No need to say, that the function is convex.
After long search I came across the notion of logarithmic derivative, elasticity and log-convex functions, though I found no definitive answer, but just a list of useless (for me) definitions.
I need a mathematically rigorous proof of my assumption or... a counterexample. All suitable functions (I know only two of such kind: $\frac{a}{bx+c}$ and $ae^{-bx+c}$ or $ae^{-bx^n+c}$ ($n>1$) ) are log-convex and satisfy the required relation. Any relevant link or a scientific paper or a reference to a math book is welcome.
If we can assume this, then $p$ is log-convex, because this is the definition for $p$ to be completely monotonic. And completely monotonic functions are log-convex by exercise 6 in section 2.1 in Convex Functions and Their Applications by Nicolescu & Persson (p. 70 here). (If you need a solution for that exercise, you might get help at Math.SE.)
If we cannot assume this, then $p(x)=\frac{\pi}{2}-\text{arctan}(x)$ seems to be a counterexample, at least for small $x$: it fits your conditions on $p$, $p'$ and $p''$, but $p''(x)p(x)-(p'(x))^2<0$ for small $x$. Here is the calculation and a plot.
A slightly more restricted version of your question with the original conditions (without complete monotonicity) could be whether $p''(x)p(x)\geq (p'(x))^2$ holds "eventually", i.e., for all $x>b$ for some $b$ (that depends on $p$).