Let $f(x,y) = g(x) h(y)$, where $g$ and $h$ both are concave and non-negative everywhere. Is $f$ also concave?
What I've tried: Let $\lambda \in [0,1]$. As $g$ is concave, $$ g(\lambda x + (1 - \lambda) x') \geq \lambda g(x) + (1 - \lambda) g(x') $$ and $h$ is concave, $$ h(\lambda y + (1 - \lambda) y') \geq \lambda h(y) + (1 - \lambda) h(y'). $$ Everything is non-negative, so we can safely multiply the two inequalities: $$ g(\lambda x + (1 - \lambda) x') \cdot h(\lambda y + (1 - \lambda) y') \geq (\lambda g(x) + (1 - \lambda) g(x')) \cdot (\lambda h(y) + (1 - \lambda) h(y')). $$ (The left-hand side is equal to $f(\lambda (x,y) + (1 - \lambda)(x',y'))$. For $f$ to be concave, this should be at least $\lambda f(x,y) + (1 - \lambda) f(x',y')$.)
Expanding the right-hand side, $$ \lambda^2 g(x) h(y) + \lambda (1 - \lambda) (g(x) h(y') + g(x') h(y)) + (1 - \lambda)^2 g(x') h(y'). $$ Rewrite using the definition of $f$: $$ \lambda^2 f(x,y) + \lambda (1 - \lambda) (f(x,y') + f(x',y)) + (1 - \lambda)^2 f(x',y'). $$ Can we argue that this is at least $\lambda f(x,y) + (1 - \lambda) f(x',y')$?