Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2

Prove that:

If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$

From Barnard & Child's "Higher Algebra".

I know that the highest power of a prime $p$ contained in $N$! is

$$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ... $$

I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n > 2$.

It is an easy exercise to show that for all real numbers $x$ we have $$ \lfloor 3x\rfloor=\lfloor x\rfloor+\lfloor x+\frac13\rfloor+\lfloor x+\frac23\rfloor. $$ Thus for all $n$ and all prime powers $p^t\ge3$ we have $$ \begin{aligned} \lfloor \frac{3n}{p^t}\rfloor&=\lfloor \frac{n}{p^t}\rfloor+ \lfloor \frac{n}{p^t}+\frac13\rfloor+\lfloor \frac{n}{p^t}+\frac23\rfloor\\ &=\lfloor \frac{n}{p^t}\rfloor+ \lfloor \frac{n+\frac{p^t}3}{p^t}\rfloor+\lfloor \frac{n+\frac{2p^t}3}{p^t}\rfloor\\ &\ge \lfloor \frac{n}{p^t}\rfloor+ \lfloor \frac{n+1}{p^t}\rfloor+\lfloor \frac{n+2}{p^t}\rfloor. \end{aligned} $$ This leaves us to deal with the case $p^t=2$. But because $n>2$, we see that $3n$ exceeds one power of two higher than any of $n,n+1,n+2$. If $n>4$ (thanks, Petite Etincelle!) we have $3n>2(n+2)$, and in the cases $n=3,4$ we have $3n>8>n+2$. This gives us a necessary extra term compensating for the deficiency at $p^t=2$.

More precisely, if $n=2k+1$ is an odd integer, then $\lfloor \dfrac{3n}2\rfloor=3k+1$ in comparison to $$ \lfloor\frac n2\rfloor+\lfloor\frac{n+1}2\rfloor+\lfloor\frac{n+2}2\rfloor=k+(k+1)+(k+1)=3k+2. $$ On the other hand, if $n=2k$ is even, then $\lfloor \dfrac{3n}2\rfloor=3k$ and $$ \lfloor\frac n2\rfloor+\lfloor\frac{n+1}2\rfloor+\lfloor\frac{n+2}2\rfloor=k+k+(k+1)=3k+1. $$ In either case we are missing a single factor two, so having that single extra term suffices.

Summing the above inequalities for $p^t\ge3$ and coupling the terms corresponding to $p^t=2$ and $p^t=2^\ell$, where $\ell$ is the largest integer such that $2^\ell\le 3n$ shows that for all primes $p$ we have $$ \sum_{t>0}\lfloor\frac{3n}{p^t}\rfloor\ge \sum_{t>0}\lfloor\frac{n}{p^t}\rfloor+\sum_{t>0}\lfloor\frac{n+1}{p^t}\rfloor+\sum_{t>0}\lfloor\frac{n+2}{p^t}\rfloor. $$ The claim follows from this.