Find the $a, b, c$ strictly positive integers such that: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
We can assume without loss of generality that $0<a \leq b \leq c$, so we have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{x}$, hence $x \leq 3$. Now we have $x=2$ or $x=3$. We treat these two cases separately and use the same method, which ultimately gives us the solution triplets $(2,3,6),(2,4,4)$, and $(3,3,3)$ (as well as the permutations $(2,6,3),(3,2,6), \ldots$ since the problem does not specify $a \leqslant b \leqslant c$).