I discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $f: \Omega \subset \mathbb{R} \rightarrow \mathbb{R\_+^\*}$ is $\alpha$-concave if $f^\alpha$ is concave.


Let $f$ a $\alpha$-concave function, $g$ a $\beta$-concave function. Let $\gamma$ be the half of the harmonic mean of $\alpha$ and $\beta$, ie. \begin{equation\*}\frac{1}{\gamma} = \frac{1}{\alpha} + \frac{1}{\beta}.\end{equation\*} Then $fg$ is a $\gamma$-concave function.


This can be proved by computing the hessian of $fg$. Have you already seen this result in the literature?



Product of concave functions and harmonic mean
$\newcommand{\a}{\alpha}$
$\newcommand{\b}{\beta}$
$\newcommand{\g}{\gamma}$


We want to prove that $ (f(ax+by)g(ax+by))^\g \ge a(f(x)g(x))^\g + b (f(y)g(y))^\g$ for every $x, y$ and $a+b = 1$. Since we know that $f(ax+by)^\a \ge af(x)^\a + bf(y)^\a$ and $g(ax+by)^\b \ge ag(x)^\b + bg(y)^\b$ it is enough to prove that


$$(af(x)^\a + bf(y)^\a)^{\frac{\g}{\a}} (ag(x)^\b + bg(y)^\b)^{\frac{\g}{\b}}\ge af(x)^\g g(x)^\g + b f(y)^\g g(y)^\g.$$


Consider set $\{x, y\}$ as measurable space with $m(x) = a$, $m(y) = b$. Then our desired inequality (after rising to the power $\frac{1}{\g}$) is nothing but Holder inequality with parameters $\a, \b$.


Unfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.


