Background :
Playing and building with Goegebra I get the inequality :
Let $1\leq x\leq \sqrt{2}$ then we have :
$$\Gamma{(x^{\Gamma{(x^{\Gamma{(x^2)}}})}-x^{\Gamma{(x)}}+x})\geq \Gamma{(x)}$$
First fact :
the function Gamma is decreasing on the interval above so we can remove one of the function Gamma.
Remains to show :
$$x^{\Gamma{(x^{\Gamma{(x^2)}}})}-x^{\Gamma{(x)}}\geq 0$$
Or :
$$\Gamma{(x^{\Gamma{(x^2)}}})-\Gamma{(x)}\geq 0$$
Or again after some steps :
$$x^{\Gamma{(x^2)}-1}\leq 1$$
Now taking the logarithm :
$$(\Gamma{(x^2)}-1)\ln(x)\leq 0$$
Wich is obvious !
Now the problem :
Claim :
Let $1\leq x\leq \sqrt{2}$ then prove or disprove :
$$x^{\Gamma{(x^{\Gamma{(x^2)}}})}-x^{\Gamma{(x)}}-(x-1)^3(x-\sqrt{2})^2\left(1-\Gamma\left(\left(\frac{\left(1+\sqrt{2}\right)}{2}\right)^{2}\right)\right)^{2}\geq 0$$
Curious fact :
The value for the minimum of the Gamma function ($x>1$) seems to be close to :
$$\Gamma\left(\left(\frac{\left(1+\sqrt{2}\right)}{2}\right)^{2}\right)$$
I cannot show this refinement alone so my question :
How to (dis)prove the claim ?
Thanks!
Let $$f(x)=\text{lhs - rhs}$$ Developinga as series, we have $$f(x)=(x-1)^3 \left(2 \gamma ^2+\left(3-2 \sqrt{2}\right) \left(\Gamma \left(\frac{1+\sqrt 2}{2} \right)^2-1\right)\right)+O\left((x-1)^4\right)$$ the coefficient being $\sim 0.638841$.
I shall not type the too long expression at the other bound; numerically $$f(x)=0.0114867 (\sqrt 2-x)+0.20949 (\sqrt 2-x)^2+O\left((x-\sqrt 2)^3\right)$$
Just to give a nice looking result (thanks to $ISC$) the first derivative apparently cancels only once close to $$x_*\sim\zeta \left(\frac{1}{2}\right) \left(\sqrt[3]{2}-\Gamma \left(\frac{5}{12}\right)\right)$$ This in an absolute error of $7.51\times 10^{-8}$; for this magic number $f'(x_*)=1.51\times 10^{-8}$, $f(x_*)=0.00181634 $ and $f''(x_*)=-0.201109$.
So $x_*$ corresponds to the maximum of the function.
Performing a numerical optimization, the maximum value of $f(x)$ is $0.00181634$.
What would remain is to prove that $f'(x)$ does not show any other root.