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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 $
$$ \Gamma(x) + \Gamma(y) \le \Gamma(x+y) \quad ?$$
$a = 2$ will do, because (letting $x \ge y$ wlg),
$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$.
Required, but never shown