Example 10 Given positive numbers $x, y, z$ satisfy $x+y+z=x y z$, and the inequality $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \leqslant \lambda$ always holds, find the range of $\lambda$.
Parse $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \leqslant \frac{1}{2 \sqrt{x y}}+\frac{1}{2 \sqrt{y z}}+\frac{1}{2 \sqrt{z x}}$ $=\frac{1}{2}\left(1 \times \sqrt{\frac{z}{x+y+z}}+1 \times \sqrt{\frac{x}{x+y+z}}+1 \times \sqrt{\frac{y}{x+y+z}}\right)$ $\leqslant \frac{1}{2}\left[\left(1^{2}+1^{2}+1^{2}\right)\left(\frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}\right)\right]^{\frac{1}{2}}$ $=\frac{\sqrt{3}}{2}$, I used substitution. The equality holds if and only if $x=y=z=\sqrt{3}$. Therefore, the range of $\lambda$ is $\left[\frac{\sqrt{3}}{2},+\infty\right)$.