About the inequality $f(-x)<\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$ Prove or disprove that $-1< x<0$ then we have :
$$f(-x)<\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$$
Where :
$$f(x)=2-x^{-\frac{212}{1000}x^{x^{\frac{1}{5}x^{2x^{\frac{1}{5}x}}}}}$$

My motivation :
To describes it I just use words . Can we approximate a curve using power towers and also somes coefficient ? In my example (if true) the two functions seems to have no link .So is it feasible in the exotic maths world ? So let's try this example !


My attempt :
I have tried basic manipulation .We can delete the coefficient $2$ wich are not in the power tower  using a new and positive coefficient (here $-\frac{212}{1000}$) because there is a symmetry with $y=1$*.Next I have tried power series but it becomes very complicated because the inequality is tight and we needs a sufficiently large ordrer wich is a negative point .
*This point seems too complicated because of the extrema .


My second attempt :
Define for $1\leq x\leq 2$:
$$p(x)=\left(\Gamma(x)\tanh(x)\right)^{\frac{2}{x(1+\tanh(0.25x))}}$$
This function seems to be convex as represented via :
$$p(x)=g(x)^{h(x)}$$
Where $g(x)$ and $h(x)$ are also convex .
So now we can use a chord with an arbitrary accuracy following the lenght of the interval wich is in itself contains in $[1,2]$.
Edit :
It seems we have on $(0.93,1)$ :
$$f(-x)<t(-x)<\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$$
Where :
$$t(x)=2-x^{-\frac{212}{1000}x^{\frac{1022}{1000}x^{\frac{1}{5}x}}}$$



Question :
How to (dis)prove it ?
Thanks in advance !
Ps:feel free to copy/paste it in Desmos .
 A: This is a very intriguing idea that I have not seen before.
I'm not too sure if this qualifies as proof, but using Python 3's Scipy & Numpy, we find that
$$I=\int_{-1}^0 h(g(x)-f(-x))dx=0\pm0$$
where $f(x)$ is your power-tower function as defined in the question, $$g(x)=\left(\Gamma(1-x)\right)^{\frac{1}{1-x}},$$ and
$$
\displaystyle   
h(x)=\frac{1}{2}\left(|x|-x\right)=\left\{ \begin{array}{ll}  
x & :\  x < 0 \\  
0 & :\ \text{otherwise.}  
\end{array} \right.$$
In other words, I is the total negative area of $g(x)-f(-x)$. If we are to assume that $f(-x)<g(x)$ for $x\in(-1,0)$, then it must be the case iff $I=0$.
The code is here:
from scipy.special import gamma
from scipy.integrate import quad
from numpy import power, divide, multiply

def f(x):
    return 2-power(x,-multiply(divide(212,1000),power(x,multiply(divide(1,5),power(x,multiply(2,power(x,multiply(divide(1,5),x))))))))

def g(x):
    return power(gamma(1-x),divide(1,1-x))

def h(x):
    return divide(abs(x)-x,2)

print(quad(lambda x: h(g(x)-f(-x)), -1, 0))

Which prints:
(0.0, 0.0)

Try it Online!
Scipy's quad method returns a value and error bounds. Thus, I am a bit suspicious of the fact that this method returned an error of exactly zero. If my code is without fault, then indeed a value of $0\pm 0$ should prove the inequality that you're looking for.
This may not be a fluke, however, as the same method finds that
$$\int_{-2}^0 h(g(x)-f(-x))dx=-0.0015678998521964117\pm 6.0928524697311665\cdot 10^{-9},$$
which is expected, as, after graphing in Desmos, it is visually clear that $f(-x)>g(x)$ for $x<-1$.
It also finds that the total area and total positive area of $g(x)-f(-x)$ for $x\in(-1,0)$ are both
$$0.2106029381090415\pm1.0430535324346124\cdot 10^{-9}.$$
Thus, in worst case scenario, if we are to assume these two (equivalent) error bounds are independent, then at the very least we can assert that
$$I=0\pm 2.0861070648692248\cdot 10^{-9}.$$
I'm not confident on the inner-workings of the quad method, but I believe this would indeed constitute as a proof if one can also prove that $f$ and $g$ are smooth on the given interval. However, that may be unnecessary.
A: $$g(x)=\Big[\Gamma (1-x)\Big]^{\frac{1}{1-x}}-f(-x)$$
At the bounds, we have
$$g(-1)=0 \qquad \text{and} \qquad g'(-1)=\frac{\gamma }{2}-\frac{36}{125} >0$$
$$g(0)=0 \qquad \text{and} \qquad g'(0)=-\infty$$
By inspection, the first derivative cancels close to $-\frac 1{20}$. A series expansion around this value predicts an extremum at $x_*=-0.053514$ while a rigorous calculation gives $x_*=-0.053807$. For this value, $h(x_*)=0.007200$ and the second derivative test confirms that this is a maximum.
Edit
Continuing with more and more approximations, the maximum is extremely close to
$$x=\frac{1}{390} \left(1-2 \sqrt{701} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{28430}{701 \sqrt{701}}\right)\right)\right)$$ (absolute error equal to $2.21\times 10^{-9}$)
