Prove that for $x\in [0,1)$

$$\Gamma(x)\ge \frac{\gamma}{x}+(1-\gamma)$$

where $\Gamma$ is the Gamma function and $\gamma$ is the Euler–Mascheroni constant.


closed as off-topic by Simply Beautiful Art, Namaste, JonMark Perry, Daniel W. Farlow, Leucippus Aug 16 '17 at 5:00

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  • 1
    $\begingroup$ Have you tried anything? As is, this is a problem statement. Please consider reading how to ask a good question and editing your post so that it is not closed. For example, it might help to multiply both sides by $x$. $\endgroup$ – Simply Beautiful Art Aug 15 '17 at 16:50
  • $\begingroup$ Yes, I wrote the inequality as $$x\Gamma(x)\ge\gamma+x(1-\gamma)$$ then I use the identity $x\Gamma(x)=\Gamma(x+1)$. Then I used this inequality valid for $x\ge 1$ $$\Gamma(x)\ge x^{(1-\gamma)x-1}$$ Then the given inequality is equivalent to prove that $$(x+1)^{(1-\gamma)(x+1)-1}\ge \gamma+x(1-\gamma)$$ and here I didn't find a way to continue. $\endgroup$ – techer Aug 15 '17 at 16:55
  • $\begingroup$ $\Gamma(1)=1$, $\Gamma(x+1)=x \Gamma(x)$, $\gamma = -\Gamma'(1)$ and a convexity argument. $\endgroup$ – reuns Aug 15 '17 at 17:14
  • $\begingroup$ The argument of convexity gives $\Gamma(1+x)\ge 1-\gamma x$, but this is not $\ge \gamma+x(1-\gamma)$. $\endgroup$ – techer Aug 15 '17 at 17:25
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    $\begingroup$ The convexity argument works, but if it is applied at $x=1$ (and not at $x=0$), since then it reads $$x\Gamma(x)=\Gamma(1+x)\geqslant\Gamma(2)+(x-1)\Gamma'(2)=1+(x-1)(1-\gamma)=\gamma+x(1-\gamma)$$ as desired. $\endgroup$ – Did Aug 15 '17 at 17:45

By the Bohr-Mollerup theorem, $\log\Gamma$ is a convex function on $\mathbb{R}^+$, hence it is a convex function on $(0,2]$. The inequality we want to prove is equivalent to $$ \Gamma(x+1) \geq \gamma +(1-\gamma)x $$ but the RHS is just the equation of the tangent line to the graph of $\Gamma(x+1)$ at $x=1$, hence such inequality is trivial by convexity.


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