Let $A,M,$ and $C$ be nonnegative integers such that $A+M+C=10$. What is the maximum value of $AMC+AM+MC+CA$? In 2000 AMC 10
  
  Let $A,M,$ and $C$ be nonnegative integers such that $A+M+C=10$. What is the maximum value of $AMC+AM+MC+CA$?


My attempt, 

$(A+1)(M+1)(C+1)=AMC+AM+AC+A+MC+M+C+1$

$=(AMC+AM+AC+MC)+11$

$AMC+AM+MC+CA=(A+1)(M+1)(C+1)-11$

I stuck at here. I went online and search for the solution in here 2000 AMC Problem, but I don't really understand for it. Hope someone can explain it for me. Thanks in advance.
You can end your solution:

By AM-GM $$AMC+AM+AC+MC=(A+1)(M+1)(C+1)-11\leq$$
$$\leq\left(\frac{3+A+M+C}{3}\right)^3-11=70.3....$$
The equality does not occur because $\{A,M,C\}\subset\mathbb N\cup\{0\}$.

Thus, $(A+1)(M+1)(C+1)\leq81$.

Let $a=A+1$, $b=M+1$ and $c=C+1$.

Hence, $a+b+c=13$ and $\{a,b,c\}\subset\mathbb N$.

We see that $abc=81$ is impossible (if $abc=81$ then $a$, $b$ and $c$ divided by $3$, which is impossible) and $abc=80$ occurs for $a=b=4$ and $b=5$.

Thus, the answer is $69$.
AMC+AM+MC+CA=(A+1)(M+1)(C+1)-11
(this is a little discovery I made)
The product or sum of any sequence of numbers is at its greatest when the numbers are more equal to each other. Sort of like you know that $(A+1)(M+1)(C+1)$ has to equal something but at the same time, $A+M+C=10$. What is the sum of 3 numbers that's equal to 10 and yet the value is closest to each other? After a few attempts, we conclude that 4,4,2 is the set. Thus,
$(4+1)(4+1)(2+1)-11 is AMC+AM+MC+CA$ 's value at its greatest.