Example 4 Let $n>1$ be a given integer. Find the largest constant $\lambda(n)$, such that for any $n$ non-zero complex numbers $z_{1}, z_{2}$, $\cdots, z_{n}$, we have
$$
\sum_{k=1}^{n}\left|z_{k}\right|^{2} \geqslant \lambda(n) \min _{1 \leq k \leq n}\left\{\left|z_{k+1}-z_{k}\right|^{2}\right\}\left(z_{n+1}=z_{1}\right) .
$$
```
\begin{array}{l}
\text { Let } \lambda_{0}(n)=\left\{\begin{array}{ll}
\frac{n}{4}, & n \text { is even; } \\
\frac{n}{4 \cos ^{2} \frac{\pi}{2 n}}, & n \text { is odd. }
\end{array}\right. \\
\text { We will prove: } \lambda_{0}(n) \text { is the required maximum constant value. } \\
\text { If there exists a positive integer } k(1 \leqslant k \leqslant n) \text {, such that } \\
\begin{aligned}
\left|z_{k+1}-z_{k}\right|=0, \\
\text { then the original inequality is obviously true. }
\end{aligned} \\
\text { Without loss of generality, assume } \\
\min _{1 \leqslant k \leqslant n}\left\{\left|z_{k+1}-z_{k}\right|^{2}\right\}=1 \text {. } \\
\text { Under this condition, we only need to prove that the minimum value of } \sum_{k=1}^{n}\left|z_{k}\right|^{2} \text { is } \lambda_{0}(n). \\
\text { First, consider the case where } n \text { is even. } \\
\frac{1}{4} \sum_{k=1}^{n} \left|z_{k+1} - z_{k}\right|^{2} \geq \frac{n}{4} \\
\text { The equality holds when } \left|z_{k+1} - z_{k}\right| = 1 \text { for all } k. \\
\text { Next, consider the case where } n \text { is odd. } \\
\left|z_{k+1} - z_{k}\right|^{2} = \left|z_{k} - z_{k+1}\right|^{2} + 2 \left|z_{k}\right| \left|z_{k+1}\right| \cos \theta_{k} \\
\text { If this is true, then by (1), we have } \\
\left|z_{k+1} - z_{k}\right|^{2} \leq \left(\left|z_{k}\right|^{2} + \left|z_{k+1}\right|^{2}\right) \left[1 + (-2 \cos \theta_{k})\right] \\
\text { We now consider two cases: } \\
\text { (1) If } \theta_{k} \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \text { for all } k, \text { then } \\
\text { Let } f(x) = \frac{1}{\sin^{2} x} \text { for } x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]. \\
\text { It is easy to see that } f(x) \text { is a convex function. } \\
\text { By (4) and Jensen's inequality, combined with (5) and (6), we get } \\
\frac{n}{4} \cdot \frac{1}{\cos^{2} \frac{\pi}{2n}} = \lambda_{0}(n). \\
\text { (2) If there exists } j (1 \leqslant j \leqslant n) \text { such that } \theta_{j} \notin \left(\frac{\pi}{2}, \frac{3\pi}{2}\right), \text { then } \\
\text { By (2), for } j \in I, \text { we have } \\
\left|z_{j}\right|^{2} + \left|z_{j+1}\right|^{2} \geq 1; \\
\text { Note that, } \\
\cos^{2} \frac{\pi}{2n} \geq \frac{n}{n+1} \\
\text { When } n = 3, \text { (9) holds; } \\
\text { When } n \geq 5, \\
\sin^{2} \frac{\pi}{2n} < \left(\frac{\pi}{2n}\right)^{2} < \frac{\pi^{2}}{2n} \cdot \frac{1}{n+1} < \frac{1}{n+1}, \\
\text { so (9) also holds. } \\
\text { Therefore, for all odd } n \geq 3, \text { (8) holds. } \\
\text { Combining (7), we have } \\
\sum \left|z_{k}\right|^{2} \geq \frac{n}{4} \cdot \frac{1}{\cos^{2} \frac{\pi}{2n}} = \lambda_{0}(n). \\
\text { In summary, the maximum value of } \lambda(n) \text { is } \lambda_{0}(n).
```