Let $A_{i}=\sqrt{x_{i} y_{i}}+z_{i}, B_{i}=\sqrt{x_{i} y_{i}}-z_{i} (i=1,2, \cdots, n)$. Using the Cauchy-Schwarz inequality and its generalization, we have:
$$\begin{aligned} & \left[\sum_{i=1}^{n} x_{i} \cdot \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}\right] \cdot \sum_{i=1}^{n} \frac{1}{x_{i} y_{i}-z_{i}^{2}} \\ \geqslant & \left[\left(\sum_{i=1}^{n} \sqrt{x_{i} y_{i}}\right)^{2}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}\right] \cdot \sum_{i=1}^{n} \frac{1}{\left(\sqrt{x_{i} y_{i}}\right)^{2}-z_{i}^{2}} \\ = & \sum_{i=1}^{n}\left(\sqrt{x_{i} y_{i}}+z_{i}\right) \cdot \sum_{i=1}^{n}\left(\sqrt{x_{i} y_{i}}-z_{i}\right) \cdot \\ & \sum_{i=1}^{n} \frac{1}{\left(\sqrt{x_{i} y_{i}}+z_{i}\right)\left(\sqrt{x_{i} y_{i}}-z_{i}\right)} \\ = & \sum_{i=1}^{n} A_{i} \cdot \sum_{i=1}^{n} B_{i} \cdot \sum_{i=1}^{n} \frac{1}{A_{i} B_{i}} \\ \geqslant & \left(\sum_{i=1}^{n} \sqrt[3]{A_{i}} \cdot \sqrt[3]{B_{i}} \cdot \sqrt[3]{\frac{1}{A_{i} B_{i}}}\right)^{3}=\left(\sum_{i=1}^{n} 1\right)^{3}=n^{3} \end{aligned}$$
Thus, the original inequality holds.

According to the proof of the generalized Cauchy-Schwarz inequality, the condition for the original inequality to hold is $A_{1}=A_{2}=\cdots=A_{n}$; $B_{1}=B_{2}=\cdots=B_{n}$; $x_{1}=x_{2}=\cdots=x_{n}$; $y_{1}=y_{2}=\cdots=y_{n}$, i.e., when $x_{1}=x_{2}=\cdots=x_{n}$; $y_{1}=y_{2}=\cdots=y_{n}$ and $z_{1}=z_{2}=\cdots=z_{n}$.