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 $
$$ \Gamma(x) + \Gamma(y) \le \Gamma(x+y) \quad ?$$
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$\begingroup$ 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? $\endgroup$– ypercubeᵀᴹFeb 25, 2011 at 17:44
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$\begingroup$ It cannot go below $a=1.4324$ the solution of $2\Gamma(x)=\Gamma(2x)$. $\endgroup$– GEdgarJul 18, 2012 at 13:56
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1 Answer
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$a = 2$ will do, because (letting $x \ge y$ wlg),
$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$.
