$(x+y)Γ(x+y)+(y+z)Γ(y+z)+(z+x)Γ(z+x)≥6$

Question

If $x,y,z>0$ and $x+y+z=3$, prove:

$$(x+y)Γ(x+y)+(y+z)Γ(y+z)+(z+x)Γ(z+x)≥6$$

I tried convexity of $f(x)=\Gamma(x)$ and Jansen's inequality. The second part of the question asking if the following inequality hold when x,y and z are positive: $$ \\(x+y)Γ(x+y)+(y+z)Γ(y+z)+(z+x)Γ(z+x)≥(x+y+z)Γ(x+y+z)$$

Question is given by Jalil Hajimir

Answer 1

$\mathbf{\text{Hint: }}$

Use $\Gamma(z+1)=z\Gamma(z)$ and proceed with Jensen's inequality with the given condition that $$x+y+z=3$$