Is $\Gamma(x+y+1)\ge\Gamma(x+1)\Gamma(y+1)$?

Question

I apologize if this has already been asked, but I could not find it. Given $x,y\ge 0$, is it true that $$\Gamma(x+y+1)\ge \Gamma(x+1)\Gamma(y+1),$$ where $\Gamma$ is the gamma function? This certainly holds for natural numbers $x,y$, as then $$\Gamma(x+y+1) = (x+y)! \ge x!y! = \Gamma(x+1)\Gamma(y+1),$$ and some plots in Mathematica seem to show that this relation also holds true for positive reals, but I have so far been unable to find a proof myself.

Answer 1

$f(x) = \log \Gamma(x + 1)$ is convex for $x \ge 0$ (compare Bohr–Mollerup theorem) with $f(0) = 0$, and therefore superadditive: $$ f(x+y) \ge f(x) + f(y) \, , $$ see for example properties of a convex function.