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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 $

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

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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? – ypercube 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

1 Answer 1

up vote 8 down vote accepted

$a = 2$ will do, because (letting $x \ge y$ wlg),

$\Gamma(x+y) \ge \Gamma(x+2) = (x+1)x \Gamma(x) \ge 6 \Gamma(x) \ge \Gamma(x) + \Gamma(x) \ge \Gamma(x) + \Gamma(y)$.

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@Moron: thanks for cleaning up the typesetting. – whuber Aug 25 '10 at 20:37
To state it explicitly perhaps one might say that if $x>2$, then $\Gamma ^{\prime}(x)>0$. – Américo Tavares Aug 25 '10 at 20:39
@whuber: You are welcome! (and welcome to this site :-)) – Aryabhata Aug 25 '10 at 20:52
@Americo: a positive derivative does not suffice to prove superadditivity. E.g., f(x) = x-10 is not superadditive for all x > 2 because 4 = f(3+3) > f(3)+f(3) = -14. – whuber Aug 26 '10 at 0:56
@Moron: Thank you! – whuber Aug 26 '10 at 0:56

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