Is the geometric mean of two non-negative, twice-differentiable concave real functions concave?
Let us work with the following definition of concave. A twice-differentiable real function $h:[0,1]\to\mathbb{R}$ is said to be concave if $h''(x) \leq 0\ $ for all $x\in [0,1].$
It is clear that the arithmetic mean $\frac{1}{2}\left( f(x) + g(x) \right)$ of two twice-differentiable, concave, non-negative $\ \left(f(x),g(x) \geq 0\ \forall x\in [0,1] \right)$ real functions $f$ and $g$ is itself concave, because $f,g$ concave $\implies f''(x) \leq 0;\ g''(x) < 0,\ \implies \left(\frac{1}{2}\left( f(x) + g(x) \right) \right)'' = \frac{1}{2}\left( \underbrace{f''(x)}_{\leq 0} + \underbrace{g''(x)}_{\leq 0} \right) \leq 0.$
However, I am having trouble proving the goemetric mean of $f$ and $g$ are concave.
$$ \left(\sqrt{fg}\ \right)'' = \left(\left(fg\right)^{1/2}\ \right)'' = \left[ \frac{1}{2} \left( fg \right)^{-1/2} (fg)' \right]' = \left[ \frac{1}{2} \left( fg \right)^{-1/2} (f'g + g'f) \right]' $$
$$ = \frac{1}{2} \left( fg \right) ^{-1/2} (f''g + 2f'g' + g''f) - \frac{1}{4} \left( fg \right)^{-3/2} (fg)' (f'g + g'f) $$
$$ = \frac{f''g + 2f'g' + g''f}{ 2\left( fg \right)^{1/2} } - \frac{ (f'g + g'f)^2}{4\left( fg \right)^{3/2} } $$
$$ = \frac{ 2f''g^2f + 4f'g'fg + 2g''f^2g - (f')^2g^2 - (g')^2f^2 - 2f'fg'g }{4\left( fg \right)^{3/2}}, $$
so it suffices to show that $2f''g^2f + 2f'g'fg + 2g''f^2g - (f')^2g^2 - (g')^2f^2 \leq 0.\ $ But I don't see how to do this. We do know that $f\geq0,\ g\geq 0,\ f''\leq0,\ $ and $g''\leq 0,\ $ which means we know that all of the terms, other than $2f'g'fg,$ are negative. But since $2f'g'fg$ could be large positive, I'm not sure where to go from here or even if the result is true.
$$$$ Note that the product of $f$ and $g$ is not necessarily concave. For example, take: $f(x) = x,\ g(x) = x.$
$$ 2f'fg'g - (f')^2 g^2 - (g')^2f^2 = - (f'g-g'f)^2 \leq 0. $$
This completes the reasoning in the question, and proves the proposition is true.