I want to prove that
For every pair of positive integers $(m,n)$, the statement $m!n!(m+n)!|(2m)!(2n)!$ holds.
which is equivalent to
For every pair of positive integers $(m,n)$, the statement $$ \sum_{i=1}^{\infty}\left\lfloor\frac m{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac n{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {m+n}{p^i}\right\rfloor \le\sum_{i=1}^{\infty}\left\lfloor\frac {2m}{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {2n}{p^i}\right\rfloor$$ holds for every prime number $p$.
I have tried setting $m=ap+c,n=bp+d$ with $a,b,c,d$ nonnegative integers and $c,d<p$. From that, the inequality becomes $$ a+\sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ b+\sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ a+b+\left\lfloor\frac{c+d}{p}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b}{p^i}+\frac {c+d}{p^{i+1}}\right\rfloor \\\le 2a+\left\lfloor\frac{2c}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}+\frac{2c}{p^{i+1}}\right\rfloor\\+ 2b+\left\lfloor\frac{2d}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2b}{p^i}+\frac{2d}{p^{i+1}}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ \left\lfloor\frac{c+d}{p}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b+\frac {c+d}{p}}{p^i}\right\rfloor \\\le \left\lfloor\frac{2c}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2a+\frac{2c}{p}}{p^i}\right\rfloor\\+ \left\lfloor\frac{2d}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2b+\frac{2d}{p}}{p^i}\right\rfloor\text.$$
After that, I tried splitting cases based on whether $2c\ge p$, whether $2d\ge p$, and whether $c+d\ge p$. The case where $2c<p$, $2d\ge p$, and $c+d\ge p$ becomes $$\sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ 1+\sum_{i=1}^{\infty}\left\lfloor\frac{a+b+1}{p^i}\right\rfloor \le \sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}\right\rfloor+ 1+\sum_{i=1}^{\infty}\left\lfloor\frac {2b+1}{p^i}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b+1}{p^i}\right\rfloor \le \sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {2b+1}{p^i}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left( \left\lfloor\frac a{p^i}\right\rfloor+ \left\lfloor\frac b{p^i}\right\rfloor+ \left\lfloor\frac{a+b+1}{p^i}\right\rfloor\right) \le \sum_{i=1}^{\infty}\left( \left\lfloor\frac {2a}{p^i}\right\rfloor+ \left\lfloor\frac {2b+1}{p^i}\right\rfloor\right)\text.$$
How to carry on from there?
