Prime inequality - why does this work?

Question

The exercise is a follows: Find all prime triplets $(p,q,r)$ so that $pq + qr + pr > pqr$. The solution: WLOG $p \leq q \leq r$. If $pq + qr + pr > pqr$, then substituting $p$ gets $3p^2 > p^3$ where we get $p=2$ considering $p$ is a prime number. Next we get an inequality $2q + 2r + qr > 2qr$. Substituting $q$ we get $q^2+4q > 2q^2$ where we get $q=2$ or $q=3$. Now substituting those values in we get that if $q=2$, then all values for $r$ give solutions and if $q=3$ then either $r=3$ or $r=5$. The answer is correct but why? Why can you substitute $p$ and $q$ like that? Or is it just a coincidence?

Answer 1

It's more or less taking a 'worst case' approach. If $p\le q\le r$, then the worst case is $p=q=r$. That reveals that if $p\ge 3$ and $q,r \ge p$ then there are no solutions. So set $p=2$ and repeat the logic for $q$.