Find all positive integers $a, b, c$ such that $ab + bc + ac > abc$. Problem statement (from Andreescu and Andrica NT: SEP): 

Find all positive integers $a, b, c$ such that $ab + bc + ac > abc$. 

The solution starts with this: 

Assume that $a \le b \le c$. If $a \ge 3$, then $ab + bc + ac \le 3abc \le abc$, a contradiction. 

I understand why this would be a contradiction, of course, but I don't understand how the inequality works, nor what is special about $a$ being greater than $3$ rather than another number. In particular, I don't understand why $ab + bc + ac \le 3bc$. For example, why is it not true that $ab + bc +ac \le 2bc$ if we assume $a \ge 2$? I see that if $a \ge 3$, then clearly $3 \le b \le c$, which might be important for understanding the solution because now the inequality flips, but I'm still confused. I've tried to understand this solution better, but I still don't understand it. 
I also see that a similar problem has been answered here: Find all prime $a, b, c$ such that $ab+bc+ac > abc$, but while it uses a similar method, it doesn't answer my question. 
 A: It is a typo.  It should say that if $a \ge 3$, we have $ab+ac+bc \le 3bc$ because $ab, ac \le bc$.  Then $3bc \le abc$ because we assumed $a \ge 3$.  Now we have $ab+ac+bc \ge abc$ which contradicts what we want to find.
A: The contradiction demonstrates that $a$ cannot be greater than or equal to $3$. Thus we know the least element of $a,b,c$ must be $2$ or less. Since $a,b,c$ are positive integers, we know $a = 1,2$. If you plug in $a=1$, you'll quickly notice a contradiction, so you can conclude $a=2$ and go from there. 
A: $ab+bc+ca>abc \rightarrow c(a+b)>ab(c-1) \rightarrow (a+b)(1-\frac{1}{c})=ab-\frac{ab}{c}$
Now, if we can find all $a,b$ such that $a+b>ab$ then we will bound the problem. 
W.L.O.G $a≥b \Rightarrow ab≥2a≥a+b$ for $b≤2$. So, we have bounded.
So, 1. For $b=1$ $a+1≥ab=a$


*For $b=2$ $a+2≥2a \rightarrow a=2$(as we took $a≥b$)


So, for $b=1$ for all $a,c>0$ $ab+bc+ca≥ac$ and for $b=2,a=2$ this becomes $4+4c>4c$, so satisfied for all $c$.
So, the triplet will be permutations of $(x,1,y);(2,2,z)$ for $x,y,z \in \mathbb N$.
