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.
