Proof: Let the left side of the inequality be $M$, and construct the pairing expression
$$N=\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x} \text {. }$$

By the Cauchy-Schwarz inequality, we have
$$M \cdot N \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2} \text {. }$$

Also, $M-N$.
$$\begin{aligned} = & \frac{1}{x y z}\left[x^{3} y^{2}+y^{3} z^{2}+z^{3} x^{2}-\left(x^{3} z^{2}+y^{3} x^{2}+z^{3} y^{2}\right) .\right. \\ = & \frac{1}{x y z}\left[\left(x^{3} y^{2}-y^{3} x^{2}\right)+\left(z^{3} x^{2}-z^{3} y^{2}\right)-\left(x^{3} z^{2}-y^{3} z^{2}\right)\right] \\ = & \frac{1}{x y z}(x-y)\left(x^{2} y^{2}+z^{3} x+z^{3} y-z^{2} x^{2}-\right. \\ & \left.z^{2} x y-z^{2} y^{2}\right) \\ = & \frac{1}{x y z}(x-y)\left[\left(x^{2} y^{2}-z^{2} x^{2}\right)-\left(z^{2} x y-\right.\right. \\ & \left.\left.z^{3} x\right)-\left(z^{2} y^{2}-z^{3} y\right)\right] \\ = & \frac{1}{x y z}(x-y)(y-z)\left(x^{2} y+x^{2} z-z^{2} x-z^{2} y\right) \\ = & \frac{1}{x y z}(x-y)(y-z)\left[\left(x^{2} y-z^{2} y\right)+\right. \\ & \left.\left(x^{2} z-z^{2} x\right)\right] \\ = & \frac{1}{x y z}(x-y)(y-z)(x-z)(x y+y z+z x) . \end{aligned}$$

Given $x \geqslant y \geqslant z>0$, we know $M-N \geqslant 0$, i.e., $M \geqslant N$.
Thus, $M^{2} \geqslant M \cdot N \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}$.
Therefore, $M \geqslant x^{2}+y^{2}+z^{2}$.