Maximization of a certain ratio for a concave positive function Assume $g : (0, 1) \to (0, +\infty)$ is a concave twice continuously differentiable function.
We want to make the ratio
$$
f(x) := -\frac{g''(x)}{g(x)}
$$
grow as $x \to 0$ as fast as possible. Some examples:
$$
g(x) = x^p,\ p \in (0, 1),\ \mbox{and}\ f(x) = \frac{p(1 - p)}{x^2}.
$$
This is maximal when $p = 1/2$, then $f(x) = 1/(4x^2)$. Also,
$$
g(x) = -x\log x\ \mbox{and}\ f(x) = -\frac1{x^2\log x},
$$
which does not grow as fast as $1/(4x^2)$ as $x \to 0$. 
 A: Sure. Just do the most natural things. 
Assume that you can have $g''<-(\frac 14+\delta)x^{-2}g(x)$ on a short interval starting at $0$ with some $\delta>0$. Note that the inequality is invariant under scalings $g(x)\mapsto ag(bx)$, so you can stretch the interval as much as you want and normalize to $g(1)=1$. Now use the compactness of concave functions normalized in such way (harnessing them with a fixed multiplicative convolution to avoid any issues with the differentiablity of the limit if you do not feel like working with generalized derivatives at the moment) and get a solution of the same differential inequality on the whole positive semi-axis. 
Since you already suspect that $\sqrt x$ is the worst you can have, write $g(x)=u(x)\sqrt x$, differentiate honestly, and arrive at the inequality
$$
u''+x^{-1}u'+\delta x^{-2}u\le 0\\,.
$$
Now rewrite it as 
$$
(xu')'+\delta x^{-1}u\le 0
$$
Since $u>0$, this means at the very least that $xu'$ is decreasing. If $u'$ gets negative anywhere, then you get $u'(x)\le -\frac cx$ at infinity. But then the integral of $u'$ diverges to $-\infty$, so $u$ gets negative itself somewhere, which is impossible. Thus $u'\ge 0$ all the way. But then $u\ge c$, so $(xu')'\le -\frac {\delta c}x$, whence (using the divergence of the harmonic integral again) $xu'$ tends to $-\infty$ making $u'$ eventually negative. Thus, "Kuda ne kin', vezde klin" (I surmise you understand Russian).
Of course, you can achieve $\frac{g''}g<-\frac{1}{4x^2}$ by considering $g(x)=\sqrt x-x^2$ on $(0,1)$ or something like that. So, strictly speaking, $\frac{1}{4x^2}$ is unbeatable only asymptotically, but I assume that this is what you meant from the beginning when asking the question.
I should confess that you arose my curiosity by claiming that you need a solution urgently and reviving an old post of mine to attract my attention. Normally you should understand that people visit MO in their free time and have no obligations whatsoever as to how fast to respond or whether to respond at all. So, I naturally wonder why it couldn't wait for a few hours or days :).
