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.