If $x,y>0,$ prove: $\quad x⁡Γ(y)+yΓ(x)\geq (x+y)Γ(\frac{x+y}{2})$

Question

If $x,y>0,$ prove: $$xΓ(y)+y⁡Γ(x)\geq (x+y)Γ\left(\frac{x+y}{2}\right)$$

I tried to solve this using the definition of the Gamma function. My attempts failed.

A plot of the function $f(x)=x\cdot\Gamma(x)$

The following inequality is also true. If $x,y>0,$ prove: $\quad x⁡Γ(x)+yΓ(y)\geq (x+y)Γ(\frac{x+y}{2})$

Question from Jalil Hajimir

Answer 1

One approach, even if I was not able to finish all the calculations at the moment.

If we put $x+y=a$, than we have to find :

$min_{0<x<a} x\Gamma(a-x)+(a-x)\Gamma(x)$

where $a$ is now a constant.

Call $f(x)=x\Gamma(a-x)+(a-x)\Gamma(x), 0<x<a$.

Than it is easy to verify that $f'(a/2)=0$. Therefore $x=a/2$ is a good candidate for a minimum. From some trials it looks that $f(x)$ is convex even if at the moment I have not been able to prove it. This would imply that the local minimum is unique and global.

Note that this is equivalent to show that:

$g(x)=x\Gamma(a(1-x))+(1-x)\Gamma(xa), 0<x<1$.

is convex, i.e. the $g''(x) \ge 0,0<x<1$, which seems to be the case from some experiments.