A function $f$ is superadditive if $f(x) + f(y) \le f(x+y)$. The question is:

Does a real number $a$ exists such that for all real numbers with $x, y\ \ge \ a$

$$\Gamma(x) + \Gamma(y) \le \Gamma(x+y) \quad ?$$

• Nice question. So, with whuber's solution, a=2 suffices to prove that it is. Now, what's the smallest a so Γ is superaddictive? – ypercubeᵀᴹ Feb 25 '11 at 17:44
• It cannot go below $a=1.4324$ the solution of $2\Gamma(x)=\Gamma(2x)$. – GEdgar Jul 18 '12 at 13:56

$a = 2$ will do, because (letting $x \ge y$ wlg),
$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$.