1

Consider the function $f(x, y) = 1 - e^{-(x + y)}$. Now $f(0, 0) = 0$, $f$ is increasing and concave, since $g(t) = -e^{-t}$ is concave. But $f(1, 0) + f(0, 1) = 2(1 - e^{-1}) > 1 - e^{-2} = f(1, 1)$, hence the claim doesn't hold even when $n = 2$. The claim doesn't hold for convex functions either. $f(x, y) = 2(x + y) + |x - y|$ is one counterexample: $...


Only top voted, non community-wiki answers of a minimum length are eligible