Problem :
Let $x_{min}=0.4616\cdots$ be the minimum of the factorial function then we have for $x,y>0$ :
$$x^2\left(x_{min}!\right)^{3}\leq \left(x!\right)^{2}\left(\frac{1}{x+y}\right)!$$
Some remarks :
Starting some trial with Desmos at first glance the minimum was near by : $e^{\frac{1}{e}}$
This remark above follows from considering the function $x^{-x}$
It's really easy to solve it without $x^2$ but I don't know how to proceed from here.
How to show it ?
We know that $x_{\min}!\le \left(\frac{1}{x+y}\right)!$ from the definition of $x_{\min}$, so it would be enough to show that $x^2(x_{\min}!)^2\le (x!)^2$. Equivalently, taking square roots, it's enough to show that $x_{\min}!\cdot x\le x!$. But we can just divide by $x$ here to get $x_{\min}!\le (x-1)!$ And this is again true by the definition of $x_{min}$*.
*: You'd actually need to have that $x_{min}!$ is the minimum for $x!$ for $x>-1$, not just for $x>0$.