2. Determine all triples $(x, y, z)$ of positive integers with $x \leq y \leq z$ that satisfy

$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=3
$$
2. If all numbers are greater than or equal to 3, then the product is less than $\left(\frac{4}{3}\right)^{3}$, which is less than 3. Therefore, the smallest of the numbers, $x$, must be 1 or 2. Case $x=1$. Then it must hold that: $\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{3}{2}$. For $y=1$, this gives $z=-4$, so no solution, for $y=2$ we get $\left(1+\frac{1}{z}\right)=1$ with no solution for $z$.

For $y>2$, the equation $\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{3}{2}$ can be rewritten as $z=2+\frac{6}{y-2}$.

For $y=3$, you find $z=8$, for $y=4$ you find $z=5$. For all larger values of $y$, $z$ becomes smaller than $y$.

Case $x=2$. Then it must hold that $\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=2$. And this can be rewritten as $z=1+\frac{2}{y-1}$. For $y=2$, you find $x=3$. For all larger values of $y$, $z$ is again smaller than $y$.

The solutions are thus $(1,3,8),(1,4,5)$ and $(2,2,3)$.