Problem :
Define :
$$f(x)=\left(e^{\frac{3x\left(\pi^{x}-e^{x}\right)}{\pi^{x}+e^{x}}}-e^{\sqrt{2x}-3}\right)^{\frac{3}{\pi}}$$
Let $x>\frac{1}{10}$ then prove or disprove that :
$$j(x)=f\left(x\right)\cdot\frac{1}{\left(f\left(0\right)\right)^{\frac{3}{\pi}}}>x!$$
A weaker result
From here ( About the inequality $\left(\ln\left(\frac{4+x}{2x+2}\right)\right)^{2}\leq x!$ ) we have (see RiverLi's answer):
$$-\ln \Gamma(x) = \ln x + \gamma x +
\sum_{n=1}^\infty [\ln(1 + x/n) – x/n],$$
We need to show :
$$\frac{3x\left(\pi^{x}-e^{x}\right)}{\pi^{x}+e^{x}}>\ln x!$$
Unfortunetaly we cannot do the same thing as in RiverLi's answer wich works because it's a lower bound and here a upper bound .
Anyway next I simplify the fraction and tried to use derivative without to reach the desired goal .
Edit :
We can play with the three function :
$$g(x)=\left(x!\right)^{\frac{1}{x}}$$
$$h(x)=x!$$
And :
$$l(x)=\left(\left(xe^{-1}+g\left(0\right)\right)+\left(-\left(\left(\frac{1}{x+a}\right)-\frac{1}{a}\right)\right)^{b}\right)^{x}$$
As good start we can choose :
$$b=1.35,a=2.2$$
Edit 2 :
It seems we have the following inequalities on $(\frac{1}{10},e)$ :
$$j(x)> \left(xe^{-1}+g\left(0\right)+\left(-\frac{1}{x+a}+\frac{1}{a}\right)^{b}\right)^{x}> x!$$
Where $a=3.198$ and $b=1$
Question :
How to show $(I)$ or find a counter-example ?
Ps: All my apologize for the previous question hope here that it's not trivial at all .
Best Answer
All this work has been done solving with (more than nasty) radicals quartic equations generated for approximations.
Starting with a small detail (already reported by @River Li in comments) : the inequality $j(x) > \Gamma(x+1)$ does not hold for very small values of $x$ close to $0$. $$\Delta=j(0) - \Gamma(1)=\left(1-\frac{1}{e^3}\right)^{3\frac{ (\pi -3)}{\pi ^2}}-1 \quad < 0$$ So, if the inequality holds, it will be for $0 < x_* < x $.
Considering the function $$h(x)=j(x)-\Gamma(x+1)$$ its series expansion around $x=0$ gives a quartic polynomial in $\sqrt x$. Expanded to $O\left(x^{5/2}\right)$, $h(x)=0$ for $\color{blue}{x_0=0.0278927}$ (the "exact" solution being $\color{red}{x_0=0.0278222}$).
Expanding $h'(x)$ as a series around $x=1$ to $O\left((x-1)^{5}\right)$ shows a maximum value of $h(x)$ for $\color{blue}{x_1=0.760950}$ (the "exact" solution being $\color{red}{x_1=0.761132}$) and $\color{blue}{h(x_1)=0.0888744}$ (the "exact" value being the same).
Expanding $h'(x)$ as a series around $x=e$ to $O\left((x-e)^{5}\right)$ shows a minimum value of $h(x)$ for $\color{blue}{x_e=2.55947}$ and $\color{blue}{h(x_e)=0.0109763}$.
Similarly, expanding $h''(x)$ around $x=\frac \pi 2$ (this is close to $\frac {x_1+x_e}2$), the inflection is predicted at $\color{blue}{x_i=1.74822}$ (the "exact" value being $\color{red}{x_i=1.74814}$)
Reusing the first series expansion leads to $\color{blue}{x_*=0.0671382}$ (the "exact" value being $\color{red}{x_*=0.0662558}$)
A numerical minimization shows that $\color{red}{x_{\text{min}}=2.55956}$ and $\color{red} {h({x_{\text{min}}})=0.0109763}$