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Given $x\in\mathbb{R}$ ,$\forall x\ge1$ seems to hold the following inequality: $$\Gamma(x)+\Gamma\left(\frac{1}{x}\right)\le\Gamma\left(1+x+\frac{1}{x}\right)$$ where the sign of equality holds only for $x=1.$ How can it be proven? Thanks.

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    $\begingroup$ Using $\Gamma(1/x)<x$ for $x>1$ may help : see Wikipedia and the Laurent expansion at $1$. Note that your relation is weak and should hold too with the addition replaced by the product at the left. $\endgroup$ – Raymond Manzoni Dec 5 '13 at 13:33
  • $\begingroup$ @RaymondManzoni: I guess it holds also replacing $\Gamma(x)$ with $x\Gamma(x)$ in the left. $\endgroup$ – Riccardo.Alestra Dec 5 '13 at 13:46
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    $\begingroup$ Well since this is $\Gamma(1+x)\cdots$ $\endgroup$ – Raymond Manzoni Dec 5 '13 at 13:47
  • $\begingroup$ Proving your inequality (for values of $x$ near $1$) doesn't appear that easy after closer examination sorry... (making this problem much more interesting! +1). Perhaps that more sophisticated methods like these proposed by Alzer in this paper will be of more value. I'll try to provide a solution tomorrow (if nobody else proposes one...). $\endgroup$ – Raymond Manzoni Dec 6 '13 at 0:42
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    $\begingroup$ @RaymondManzoni: The inequality seems to have been proven by Donald Kershaw in a private communication, but I'm not able to find it. $\endgroup$ – Riccardo.Alestra Dec 6 '13 at 9:42
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A proof of this (rather non-trivial for $x\approx 1$) inequality was provided by Jameson in $2012$ in 'An inequality for the gamma function conjectured by D. Kershaw'.
Concerning the product a proof is in Giordano and Laforgia's paper.

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