# Show that $\Gamma(x+1)\ge\gamma+(1-\gamma)x$ for every $x\in [0,1)$ [closed]

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, LeucippusAug 16 '17 at 5:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Simply Beautiful Art, Namaste, JonMark Perry, Daniel W. Farlow, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $x$. – Simply Beautiful Art Aug 15 '17 at 16:50
• 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. – techer Aug 15 '17 at 16:55
• $\Gamma(1)=1$, $\Gamma(x+1)=x \Gamma(x)$, $\gamma = -\Gamma'(1)$ and a convexity argument. – reuns Aug 15 '17 at 17:14
• The argument of convexity gives $\Gamma(1+x)\ge 1-\gamma x$, but this is not $\ge \gamma+x(1-\gamma)$. – techer Aug 15 '17 at 17:25
• 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. – 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.