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I've encountered the following inequality which I am not able to prove but pretty certain that it is true: $$\Gamma\Big(x+\frac{1}{2}\Big)^2 < x\Gamma(x)^2$$ This should be true for $x\in\mathbb{R}^+$. However, I only need this for $x\in\mathbb{Q}^{+}$ or even $x=\frac{n}{4}$ for $n\in\mathbb{N}$, so I can prove $$\Gamma\Big(\frac{k+7}{4}\Big)^2 < \frac{k+5}{4}\Gamma\Big(\frac{k+5}{4}\Big)^2$$ and similar expressions for $k\in\mathbb{N}$.

In general, I have trouble with terms involving $\Gamma\Big(x+\frac{1}{2}\Big)$. I am aware of Gautschi's Inequality, but find it not quite sharp enough for small $x$. So I would also be thankful for any information on inequalities for $\Gamma\Big(x+\frac{1}{2}\Big)$.

Thanks, Nils

Nils
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    Just use the functional equation for $\Gamma$ and log-convexity. – user10354138 Sep 21 '23 at 11:44
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    Gautschi’s Inequality works just fine. Take $s=1/2$, then $$ x^{1/2} < \frac{{\Gamma (x + 1)}}{{\Gamma (x + 1/2)}} = \frac{{x\Gamma (x)}}{{\Gamma (x + 1/2)}} $$ for $x>0$. Re-arrange to get $$ \Gamma (x + 1/2) < x^{1/2} \Gamma (x) $$ for $x>0$. The inequality cannot be true for all real $x$ due to the poles of the gamma function. – Gary Sep 21 '23 at 11:53
  • @Gary Thanks! And I've also meant $x\in\mathbb{R}^+$ and edited it. – Nils Sep 21 '23 at 12:49

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Gautschi's Inequality works just fine. Take $s=1/2$, then $$x^{1/2} < \frac{{\Gamma (x + 1)}}{{\Gamma (x + 1/2)}} = \frac{{x\Gamma (x)}}{{\Gamma (x + 1/2)}}$$ for $x>0$. Re-arrange to get $$\Gamma (x + 1/2) < x^{1/2} \Gamma (x)$$ for $x>0$.

Gary
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