We have $x\le y\le z\Longrightarrow \frac{1}{x}\ge \frac{1}{y}\ge \frac{1}{z}.$ Thus $3\le \left(1+\frac{1}{x}\right)^3,$ which leads to either $x=1$ or $x=2.$ 

Now find the value of $y,z$ by using this.