Prove that $\Gamma(\operatorname{W}(x))$ is convex $\forall x>0$ Background :
At the begining I was studing a function wich increases slowly and maybe have some property useful in number theory .Particulary I have found :


Let $0<x\,$ define the function :
$$f(x)=\Gamma(\operatorname{W}(x))$$


Where we see the Gamma function and the Lambert's function
Then prove that :
$$f''(x)>0\quad\forall x>0$$
Well working with WA wich is a little bit capricious I find that the minimum of the second derivative occurs on $I=[24800,24900]$
I have tried to solve the following expression see here without success .
My second strategy is : if we know that mid-point convexity and conitinuity implies convexity we can says that we have :
Let $ x,y>0$  then we have :
$$f(x)+f(y)\geq 2f\Big(\frac{x+y}{2}\Big)$$
I can solve it for large value but not on $I$ describe above .
Update :
Following the good start by TheSimpliFire we have to prove :
$$\psi(x)+\frac{(\psi(x))'}{\psi(x)}>1+\frac{1}{x+1}\quad \forall x>0$$
From the source we have (see (51) and (52)):
$$\frac{\pi^2}{\pi^2x+6-\pi^2}\leq(\psi(x))' \quad \forall x\geq 1$$
And
$$\log\Big((t-1)\frac{\pi^2}{6}+1\Big)-\gamma\leq\psi(t)<\log(2t-1)-\gamma\quad \forall t\geq 1$$
Perhaps there is an issue now .
So if you have an idea or an approach like a hint it would be nice .
Thanks a lot for all your contributions !
Max.
Source : https://www.hindawi.com/journals/jam/2014/264652/
 A: Let $f(x)=\Gamma(W(x))$ so $f'(x)=\Gamma'(W(x))W'(x)$ and $f''(x)=\Gamma''(W(x))W'(x)^2+\Gamma'(W(x))W''(x)$.
Since $W'(x)=W(x)/[x(W(x)+1)]$ and $W''(x)=-W(x)^2(W(x)+2)/[x^2(W(x)+1)^3]$ it follows that $$f''(x)=\Gamma''(W(x))\cdot\frac{W(x)^2(W(x)+2)}{x^2(W(x)+1)^2}-\Gamma'(W(x))\cdot\frac{W(x)^2}{x^2(W(x)+1)^3}$$ and convexity is equivalent to the inequality $\Gamma''(W(x))(W(x)+1)>\Gamma'(W(x))(W(x)+2)$. This simplifies to showing that $(x+1)\Gamma''(x)>(x+2)\Gamma'(x)$ for $x>0$ since $W(x)$ is bijective on the positive reals.
Since $\Gamma'(x)=\Gamma(x)\psi(x)$ and $\Gamma''(x)=\Gamma(x)[\psi(x)^2+\psi'(x)]$ where $\psi(x)=\psi^{(0)}(x)$, it suffices to show that $$(x+1)\psi(x)>(x+2)(\psi(x)^2+\psi'(x))\iff\psi(x)+\frac{\psi'(x)}{\psi(x)}>1+\frac1{x+1}$$ for $x>0$. It is known that $\psi'(x)>2\log x-2\psi(x)$[1] and $(2x)^{-1}<\log x-\psi(x)<x^{-1}$[2] for $x>1$, and so it is sufficient that $$\log x+\frac2{1-(2x\log x)^{-1}}>3+\frac1x+\frac1{x+1}$$ which is true for $x>7/2$. Hence $f''(x)>0$ for $x>(7/2)e^{7/2}$ and convexity on $x>0$ can be shown by plotting $f''(x)$ in the interval $0<x\le(7/2)e^{7/2}$.

 References 
 [1] Farhangdoost, M. R., Dolatabadi, M. K. (2014). New Inequalities for Gamma and Digamma Functions. Article ID 264652. 
 [2] Alzer, H. (1997). On some inequalities for the gamma and psi functions. Mathematics of computation. 66(217):373-389. 
