China Team Selection Test

Let $\lambda_0(n) = \begin{cases} \frac{n}{4}, & 2 \mid n, \\ \frac{n}{4 \cos^2 \frac{\pi}{2n}}, & \text{otherwise.} \end{cases}$. We prove $\lambda_0(n)$ is the greatest constant.

If there exists $k$ ($1 \leq k \leq n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds obviously. So without loss of generality, we can assume
$$\min_{1 \leq k \leq n} \{|z_{k+1} - z_k|^2\} = 1. \qquad \textcircled{1}$$
Under this condition, it is sufficient to show that the minimum value of $\sum_{k=1}^{n} |z_k|^2$ is $\lambda_0(n)$.

When $n$ is even. Since
$$\begin{aligned} \sum_{k=1}^{n} |z_k|^2 &= \frac{1}{2} \sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \ &\geq \frac{1}{4} \sum_{k=1}^{n} |z_{k+1} - z_k|^2 \ &\geq \frac{n}{4} \min_{1 \leq k \leq n} \{|z_{k+1} - z_k|^2\} = \frac{n}{4}, \end{aligned}$$
and equality holds when $(z_1, z_2, \dots, z_n) = (\frac{1}{2}, -\frac{1}{2}, \dots, \frac{1}{2}, -\frac{1}{2})$, thus the minimum value of $\sum_{k=1}^{n} |z_k|^2$ is $\frac{n}{4} = \lambda_0(n)$.

Next consider the condition when $n$ is odd. Let
$$\theta_k = \arg \frac{z_{k+1}}{z_k} \in [0, 2\pi), \quad k = 1, 2, \dots, n.$$
For all $k$, $k = 1, 2, \dots, n$. If $\theta_k \leq \frac{\pi}{2}$ or $\theta_k \geq \frac{3\pi}{2}$, then by ①,
$$|z_k|^2 + |z_{k+1}|^2 = |z_k - z_{k+1}|^2 + 2|z_k||z_{k+1}| \cos \theta_k \geq |z_k - z_{k+1}|^2 \geq 1. \quad ②$$
If $\theta_k \in (\frac{\pi}{2}, \frac{3\pi}{2})$, then since $\cos \theta_k < 0$ and by ①,
$$\begin{aligned} 1 &\leq |z_k - z_{k+1}|^2 \\ &= |z_k|^2 + |z_{k+1}|^2 - 2 |z_k||z_{k+1}| \cos \theta_k \\ &\leq (|z_k|^2 + |z_{k+1}|^2)(1 + (-2 \cos \theta_k)) \\ &= (|z_k|^2 + |z_{k+1}|^2) \cdot 2 \sin^2 \frac{\theta_k}{2}. \end{aligned}$$
Therefore
$$|z_k|^2 + |z_{k+1}|^2 \geq \frac{1}{2 \sin^2 \frac{\theta_k}{2}}. \quad ③$$

(1) If for all $k$ ($1 \leq k \leq n$), $\theta_k \in (\frac{\pi}{2}, \frac{3\pi}{2})$, by ③
$$\sum_{k=1}^{n} |z_k|^2 = \frac{1}{2} \sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \geq \frac{1}{4} \sum_{k=1}^{n} \frac{1}{\sin^2 \frac{\theta_k}{2}}. \quad ④$$
Since $\prod_{k=1}^{n} \frac{z_{k+1}}{z_k} = \frac{z_{n+1}}{z_1} = 1$, so
$$\sum_{k=1}^{n} \theta_k = \arg\left(\prod_{k=1}^{n} \frac{z_{k+1}}{z_k}\right) + 2m\pi = 2m\pi, \quad ⑤$$
where $m$ is a positive integer, and $m < n$. Notice that $n$ is odd, so
$$0 < \sin \frac{m\pi}{n} \leq \sin \frac{(n-1)\pi}{2n} = \cos \frac{\pi}{2n}. \quad ⑥$$
Let $f(x) = \frac{1}{\sin^2 x}$, $x \in [\frac{\pi}{4}, \frac{3\pi}{4}]$, it's easy to show that $f(x)$ is a convex function. By ④ and Jensen's inequality, and combining ⑤ and ⑥, we have
$$\begin{aligned} \sum_{k=1}^{n} |z_k|^2 &\geq \frac{1}{4} \sum_{k=1}^{n} \frac{1}{\sin^2 \frac{\theta_k}{2}} \\ &\geq \frac{n}{4} \cdot \frac{1}{\sin^2 \left( \frac{1}{n} \sum_{k=1}^{n} \frac{\theta_k}{2} \right)} \\ &= \frac{n}{4} \cdot \frac{1}{\sin^2 \frac{m\pi}{n}} \geq \frac{n}{4} \cdot \frac{1}{\cos^2 \frac{\pi}{2n}} = \lambda_0(n). \end{aligned}$$

(2) If there exists $j$ ($1 \leq j \leq n$), such that $\theta_j \notin (\frac{\pi}{2}, \frac{3\pi}{2}]$, let
$$I = \left\{ j \mid \theta_j \notin \left( \frac{\pi}{2}, \frac{3\pi}{2} \right),\ j = 1, 2, \dots, n \right\}.$$
By ②, for $j \in I$, we have $|z_j|^2 + |z_{j+1}|^2 \geq 1$; and by ③, for $j \notin I$, we have
$$|z_j|^2 + |z_{j+1}|^2 \geq \frac{1}{2 \sin^2 \frac{\theta_j}{2}} \geq \frac{1}{2}.$$
Therefore,
$$\begin{aligned} \sum_{k=1}^{n} |z_k|^2 &= \frac{1}{2} \left( \sum_{j \in I} (|z_j|^2 + |z_{j+1}|^2) + \sum_{j \notin I} (|z_j|^2 + |z_{j+1}|^2) \right) \\ &\geq \frac{1}{2} |I| + \frac{1}{4} (n - |I|) = \frac{1}{4} (n + |I|) \geq \frac{n+1}{4}. \end{aligned}$$

$\frac{n+1}{4} \geq \frac{n}{4} \cdot \frac{1}{\cos^2 \frac{\pi}{2n}} \Leftrightarrow \cos^2 \frac{\pi}{2n} \geq \frac{n}{n+1}$
$$\Leftrightarrow \sin^2 \frac{\pi}{2n} = 1 - \cos^2 \frac{\pi}{2n} \leq 1 - \frac{n}{n+1} = \frac{1}{n+1}.$$
The equality holds when $n = 3$; when $n \geq 5$,
$$\sin^2 \frac{\pi}{2n} < \left( \frac{\pi}{2n} \right)^2 < \frac{\pi^2}{2n} \cdot \frac{1}{n+1} < \frac{1}{n+1},$$
so the inequality also holds. So for odd integer $n \geq 3$, $\frac{n+1}{4} \geq \frac{n}{4} \cdot \frac{1}{\cos^2 \frac{\pi}{2n}}$.

On the other hand, when $z_k = \frac{1}{2\cos\frac{\pi}{2n}} \cdot e^{\frac{i(n-1)k\pi}{n}}$, $k = 1, 2, \dots, n$, we have $|z_k - z_{k+1}| = 1$, $k = 1, 2, \dots, n$, and $\sum_{k=1}^n |z_k|^2$ achieves its minimum value $\lambda_0(n)$.

In a word, the greatest $\lambda(n)$ is
$$\lambda_0(n) = \begin{cases} \frac{n}{4}, & 2 \mid n, \\ \frac{n}{4 \cos^2 \frac{\pi}{2n}}, & \text{otherwise.} \end{cases}$$