Concavity of function $x \mapsto \log \Gamma(x+1) / x$

Question

I'm looking for a proof that the function $\mathbb{R}_+ \to \mathbb{R}$, $$x \mapsto \frac {\log\Gamma(x+1)} {x}$$ is concave. The reason why I'm interested in this is because it implies inequalities like $$ (a!)^{1/a} (b!)^{1/b} \leq \left(\frac{a+b}{2}!\right)^{\frac{2}{a+b} \cdot 2}.$$

Answer 1

This is not a formal proof.

Using Stirling approximation, we have $$\frac{\log (\Gamma (x+1))}x=(\log (x)-1)+\frac{\log (2 \pi )+\log \left(x\right)}{2 x}+\frac{1}{12 x^2}-\frac{1}{360 x^4}+O\left(\frac{1}{x^{9/2}}\right)$$ So, the second derivative $$\left(\frac{\log (\Gamma (x+1))}x\right)''=-\frac{1}{x^2}+\frac{\log \left(x\right)+\log (2 \pi )-\frac{3}{2}}{x^3}+\frac{1}{2 x^4}+O\left(\frac{1}{x^6}\right)\tag 1$$ which is always negative.

Otherwise, more formally $$\left(\frac{\log (\Gamma (x+1))}x\right)''=\frac{2 \log (\Gamma (x+1))}{x^3}-\frac{2 \psi ^{(0)}(x+1)}{x^2}+\frac{\psi ^{(1)}(x+1)}{x}\tag 2$$