(UK Olympiades) Find all positive integers $x, y, z$ such that

$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=2
$$
Without loss of generality, we assume $x \leq y \leq z$. If $x \geq 4$, then $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) \leq\left(\frac{5}{4}\right)^{3}<2$. Therefore, $x=1,2$ or 3. If $x=1$, then $\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=1$ which is impossible. By studying the other cases, we find the solutions $(2,6,7),(2,5,9),(2,4,15),(3,3,8)$ and $(3,4,5)$.