Prove: By the local Cauchy inequality, we have: $\left(x^{2}+y+z\right)$ $(1+y+z) \geqslant(x+y+z)^{2}$. Therefore, $\frac{1}{\sqrt{x^{2}+y+z}} \leqslant$ $\frac{\sqrt{1+y+z}}{x+y+z}$, and similarly, we can obtain the other two inequalities. Also, by the Cauchy inequality $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+z)^{2}$ and $x^{2}+y^{2}$ $+z^{2}=3$, we know: $x^{2}+y^{2}+z^{2} \geqslant x+y+z$. Therefore, we only need to prove

Again, by the Cauchy inequality, we have $x \sqrt{1+y+z}+$
$$\begin{array}{l} y \sqrt{1+z+x}+z \sqrt{1+x+y} \leqslant \sqrt{x+y+z} \\ \sqrt{x+y+z+2 x y+2 x z+2 y z} \leqslant \\ \sqrt{x+y+z} \sqrt{x^{2}+y^{2}+z^{2}+2 x y+2 x z+2 y z}= \\ \sqrt{(x+y+z)^{3}}, \text { i.e., } \\ \frac{x \sqrt{1+y+z}+y \sqrt{1+z+x}+z \sqrt{1+x+y}}{x+y+z} \leqslant \\ \sqrt{x+y+z} \leqslant \sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{3} \end{array}$$