TITLE: An uncanny inequality with Gamma function
QUESTION [6 upvotes]: Prove for $x>0$ that
$$ \frac{\Gamma^{\prime}(x+1)}{\Gamma(x+1)}>\log x$$
How to prove this inequality? thanks. This is a problem from Miklos Schweitzer Competition.

REPLY [3 votes]: Since $\log(\Gamma(x))$ is convex, we have
$$
\frac{\log(\Gamma(x+1))-\log(\Gamma(x))}{(x+1)-x}\le\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x+1))\le\frac{\log(\Gamma(x+2))-\log(\Gamma(x+1))}{(x+2)-(x+1)}
$$
which can be rewritten, using $\Gamma(x+1)=x\Gamma(x)$, as
$$
\log(x)\le\frac{\Gamma'(x+1)}{\Gamma(x+1)}\le\log(x+1)
$$