Prove that the function $f(x)=\frac{x}{\sqrt{1-x}}(0<x<1)$ is a convex function. For $t_{1}, t_{2}>1$, and $x_{i}=1-\frac{1}{t_{i}^{2}}(i=1,2)$. Since
$$\frac{1}{2}\left(t_{1}+t_{2}\right) \geqslant \sqrt{t_{1} t_{2}}, \sqrt{t_{1}^{2}+t_{2}^{2}} \geqslant \sqrt{2 t_{1} t_{2}}$$

Thus, $\frac{1}{2}\left(t_{1}+t_{2}\right) \geqslant \frac{\sqrt{2} t_{1} t_{2}}{\sqrt{t_{1}^{2}+t_{2}^{2}}}$,
which means $\frac{1}{2}\left(\frac{1}{\sqrt{1-x_{1}}}+\frac{1}{\sqrt{1-x_{2}}}\right) \geqslant \frac{1}{\sqrt{1-\frac{1}{2}\left(x_{1}+x_{2}\right)}}$.
Also, $\left(x_{1}-x_{2}\right)\left(\frac{1}{\sqrt{1-x_{1}}}-\frac{1}{\sqrt{1-x_{2}}}\right) \geqslant 0$, hence,
$$\frac{x_{1}}{\sqrt{1-x_{1}}}+\frac{x_{2}}{\sqrt{1-x_{2}}} \geqslant \frac{1}{2}\left(\frac{1}{\sqrt{1-x_{1}}}+\frac{1}{\sqrt{1-x_{2}}}\right)\left(x_{1}+x_{2}\right),$$

Using (1), from (2) we get
$$\frac{1}{2}\left(\frac{x_{1}}{\sqrt{1-x_{1}}}+\frac{x_{2}}{\sqrt{1-x_{2}}}\right) \geqslant \frac{\frac{1}{2}\left(x_{1}+x_{2}\right)}{\sqrt{1-\frac{1}{2}\left(x_{1}+x_{2}\right)}}$$

Equality holds if and only if $x_{1}=x_{2}$. This shows that the function $f(x)=\frac{x}{\sqrt{1-x}}(0<x<1)$ is a convex function. Since $f(x)=\sqrt{x}(x>0)$ is also a convex function, we have
$$\frac{1}{n}\left(\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}\right) \leqslant \sqrt{\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}}=\frac{\sqrt{n}}{n} .$$

From (3) and (4), we obtain the desired inequality.