I wonder which mathematical construct can be used to solve inequality of factorials. I'm studying computer science and I have always used numerical computing software as Wolframalpha to solve this.
Here is an example:
$$(365-n)! \geq 2,\quad n\in [1,365]$$
On
In this case, all you need to know is that the factorial is an increasing function for non-negative integers (i.e m>n implies that m!>n!).
So, since $2!\geq2$, we have that $k!\geq 2$ whenever $k\geq 2$. However, in your example, when n=365 or 364 we have $$(365-365)! = 0! = 1$$ $$(365-364)! = 1! = 1$$ So the inequality does not hold in the whole range. It does hold only for $n \in [1,363]$
If you want $$m!\ge 2$$ then you need $m\ge 2$, since $2!=2$ and factorial is an increasing function.
Therefore you have the inequality $$365-n\ge2 \implies 1\le n\le 363$$
If there had been another number instead of $2$, say $r$ that is, if you have $$(f(n))!\ge r$$ Then you need to find a number $m$ such that $m!\ge r$ and then solve the inequality $$f(n) \ge m$$