Is it true that
For every $a,b,x,y\in\mathbb N$,
If $x\leq y$ and $ab\leq y$ and $a\leq x!$ and $b\leq (y-x)!$,
Then, $ab\leq a \sqrt {\frac{y!}{x!}}+b \sqrt {\frac{y!}{(y-x)!}}$?
My efforts so far:
- I made a Python code and assured that it is true for $a,b,x\leq 30$ and $y\leq 100$
- I managed to prove a very limited special case
By AM-GM
$$a \sqrt {\frac{y!}{x!}}+b \sqrt {\frac{y!}{(y-x)!}}\ge 2\sqrt{ab}\sqrt[4]{y!\binom y x}\ge 2\sqrt{ab}\sqrt[4]{(ab)!}\ge ab$$
since for any $n\in \mathbb N$ we have $$n!\ge \frac1{16}n^2$$