Prove that $\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$ is an integer.

Prove that $$\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$$ is an integer for all pairs of positive integers $a, b$ (American Mathematical Monthly)

My work -

$ v_{p}((3 a+3 b) !(2 a) !(3 b) !(2 b) !)=\sum_{k \geq 1}\left(\left\lfloor\frac{3 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 a}{p^{k}}\right\rfloor+\left\lfloor\frac{3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 b}{p^{k}}\right\rfloor\right) $

and

$ \begin{array}{l} v_{p}\left((2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}\right) \\ \quad \quad=\sum_{k \geq 1}\left(\left\lfloor\frac{2 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+2 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+b}{p^{k}}\right\rfloor+\left\lfloor\frac{a}{p^{k}}\right\rfloor+2\left\lfloor\frac{b}{p^{k}}\right\rfloor\right) \end{array} $

now

With the substitution $x=\frac{a}{p^{k}}, y=\frac{b}{p^{k}},$ we have to prove that for any nonnegative real numbers $x, y$ we have $\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor$

I tried putting $\{x\}+\lfloor x\rfloor=x$ and $\{y\}+\lfloor y\rfloor=y$ and i get things in terms of fractional parts but i am not able to prove after that ....

thankyou

The left-hand side minus the right-hand side of your inequality is always 0, 1, or 2. The Mathematica plot below mostly proves it. One can cut the work in half by noticing a symmetry $(x, y) \mapsto (-x, -y)$ which induces $L-R \mapsto 2-(L-R)$. The picture strongly suggests that case-by-case reasoning will be messy no matter what. Having the computer check non-negativity for a certain finite collection of points would also be a proof.

By your work we need to prove that: $$\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor,$$ where $\{x,y\}\subset[0,1).$

Let $x\in\left[\frac{i}{6},\frac{i+1}{6}\right)$, where $i$ is an integer, $0\leq i\leq 5$ and $y\in\left[\frac{j}{6},\frac{j+1}{6}\right)$, where $j$ is an integer, $0\leq j\leq 5$.

Thus, after considering these $36$ cases (maybe a bit of less) we obtain that our inequality is true.

For example, let $x\in\left[\frac{1}{3},\frac{1}{2}\right)$ and $y\in\left[\frac{1}{2},\frac{2}{3}\right).$

Thus, $$2.5\leq3x+3y<3.5,$$ $$\frac{2}{3}\leq2x<1,$$ $$\frac{3}{2}\leq3y<2,$$

$$1\leq2y<\frac{4}{3},$$ $$\frac{13}{6}\leq2x+3y<3,$$ $$\frac{4}{3}\leq x+2y<\frac{11}{6},$$ $$\frac{5}{6}\leq x+y<\frac{7}{6}$$ and $$[x]=[y]=0.$$ Id est, it's enough to prove that $$[3x+3y]+1+1\geq2+1+[x+y],$$ which is true because $$[3x+3y]\geq2$$ and $$[x+y]\leq1.$$

The rest is the similar.