Find all triplets of strictly positive integers $(x, y, z)$ with

$x \leq y \leq z$, such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
Given $\frac{1}{x} \geq \frac{1}{y} \geq \frac{1}{z}$, we must necessarily have $\frac{1}{x} \geq \frac{1}{3}$, which means $x \leq 3$. If $x=3, \frac{1}{y}+\frac{1}{z}=\frac{2}{3}$, so $y \leq 3$: the only possible solution is $(3,3,3)$. If $x=2, \frac{1}{y}+\frac{1}{z}=\frac{1}{2}$, so $y \leq 4$: there are two solutions, $(2,4,4)$ and $(2,3,6)$. And it is clear that $x=1$ does not work, because $\frac{1}{y}+\frac{1}{z}$ cannot be zero.