Proof: From $(*)$ and $r_{a}=\frac{r s}{s-a}$, etc., we know that the original inequality is equivalent to
$$\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}$$
$$\leqslant \sqrt{2}(x+y+z) \Leftrightarrow \sqrt{2 x(y+z)}+\sqrt{2 y(z+x)} \\ +\sqrt{2 z(x+y)} \leqslant 2(x+y+z).$$

According to the two-variable arithmetic-geometric mean inequality, we have
$$\begin{array}{l} \quad \sqrt{2 x(y+z)}+\sqrt{2 y(z+x)}+ \\ \sqrt{2 z(x+y)} \leqslant \frac{1}{2}[(2 x+y+z)+(2 y+z+x) \\ +(2 z+x+y)]=2(x+y+z). \end{array}$$