$m$,$n$,$p$ are natural numbers.
$2mnp=(m+2)(n+2)(p+2)$
$m\leq n\leq p$
then, find the maximum value of $p$.
the source of this question is 1998 AIME.
but the official solution is too long and not easy to get for me.
so I'm searching another solutions.
Brute force will work.
We rewrite it as $$2=\left(1+\frac{2}{m}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{2}{p}\right)$$
Now $2>1+2/m > \sqrt[3]{2}> 1 + 1/4$. Thus $2<m<8$.
You can then do case-by-case analysis. For example, if $m=3$, then you are trying to solve:
$$\frac{6}{5} = \left(1+\frac{2}{n}\right)\left(1+\frac{2}{p}\right)$$
Again, since $6/5 > 1+2/n > \sqrt{6/5}$, you can reduce to trial of finitely many $n$ - specifically, $10<n<21$.
That probably is going to take a fair amount of time to check.
My answer is p=130.
There are 4 cases to consider:
1) $m=1$. Solving for $p$ we get $$ p =-\frac{6\,n+12}{n+6} $$ which is always negative. So $m \neq 1$.
2) $m=2$. Solving for $p$ we get $$ p = -(n+2) < 0$$ So $m \neq 2$.
3) $m=3$. Solving for $p$ we get $$ p=\frac{10 n+20}{n-10} = 10 + \frac{120}{n-10}$$ So the largest value of $p$ is when $n=11$. This gives $m=3$, $n=11$, $p=130$
4) $m=4$. Solving for $p$ $$ p = \frac{6 n+12}{n-6} = 6 + \frac{48}{n-6}$$ and the maximum value for $p$ is when $n=7$ and $p=54$
It is not very hard to show that as $m$ increases, the maximum value of $p$ decreases, so the answer is $$ m=3,~ n=11, ~ p=130$$
To show that we don't have to look for $m$ beyond $m=4$ establish the following:
1) Solve for $p$ and show that $\frac{dp}{dn} <0$. So we want $n$ as small as possible.
2) Set $n=m$, the smallest value of $n$ and simplify and show that solution to $p$ monotonically decreases for $m$.
I am sure there are simpler ways to show that $m \le 4$