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
 A: 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.

A: 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.
