Prove that first transform the left side of the inequality into an average form.
$$\begin{aligned} \text { Left }= & \frac{x+y+z+\sqrt{x^{2}+y^{2}+z^{2}}}{\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} \\ = & \frac{x+y+z}{\sqrt{x^{2}+y^{2}+z^{2}}} \cdot \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} \\ & +\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} . \\ \text { Let } \lambda= & \frac{x+y+z}{\sqrt{x^{2}+y^{2}+z^{2}}}, \mu=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}, \end{aligned}$$

By Corollary 4, we have
$$\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leqslant \frac{x+y+z}{3} \leqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}$$

Therefore,
$$\lambda \leqslant \sqrt{3}, \mu \leqslant \frac{1}{3 \sqrt{3}},$$

Thus, the left side $=\lambda \mu+\mu \leqslant \frac{3+\sqrt{3}}{9}$.
In general, if $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, then
$$\frac{\sum_{i=1}^{n} a_{i}+\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{\frac{1}{2}}}{\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} \frac{1}{a^{i}}\right)} \leqslant \frac{1}{n^{2}}(n+\sqrt{n})$$