China Team Selection Test

Let ${\lambda }_0(n)=$. We prove ${\lambda }_0(n)$ is the greatest constant.

If there exists $k$ ($1\le k\le n$) such that $|z_{k+1}-z_k|=0$, the inequality holds obviously. So without loss of generality, we can assume $$\underset{1\le k\le n}{\min }\{|z_{k+1}-z_k{|}^2\}=1.\stackrel{◯}{1}$$ 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 $$\begin{array}{rlrl}\sum _{k=1}^n|z_k{|}^2& =\frac{1}{2}\sum _{k=1}^n(|z_k{|}^2+|z_{k+1}{|}^2)& \ge \frac{1}{4}\sum _{k=1}^n|z_{k+1}-z_k{|}^2& \ge \frac{n}{4}\underset{1\le k\le n}{\min }\{|z_{k+1}-z_k{|}^2\}=\frac{n}{4},\end{array}$$ 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 $${\theta }_k=\arg \frac{z_{k+1}}{z_k}\in [0,2\pi ),k=1,2,\dots ,n.$$ For all $k$, $k=1,2,\dots ,n$. If ${\theta }_k\le \frac{\pi }{2}$ or ${\theta }_k\ge \frac{3\pi }{2}$, then by ①, $$|z_k{|}^2+|z_{k+1}{|}^2=|z_k-z_{k+1}{|}^2+2|z_k||z_{k+1}|\cos {\theta }_k\ge |z_k-z_{k+1}{|}^2\ge 1.②$$ If ${\theta }_k\in (\frac{\pi }{2},\frac{3\pi }{2})$, then since $\cos {\theta }_k<0$ and by ①, $$\begin{array}{rl}1& \le |z_k-z_{k+1}{|}^2\\ & =|z_k{|}^2+|z_{k+1}{|}^2-2|z_k||z_{k+1}|\cos {\theta }_k\\ & \le (|z_k{|}^2+|z_{k+1}{|}^2)(1+(-2\cos {\theta }_k))\\ & =(|z_k{|}^2+|z_{k+1}{|}^2)\cdot 2{\sin }^2\frac{{\theta }_k}{2}.\end{array}$$ Therefore $$|z_k{|}^2+|z_{k+1}{|}^2\ge \frac{1}{2{\sin }^2\frac{{\theta }_k}{2}}.③$$

(1) If for all $k$ ($1\le k\le n$), ${\theta }_k\in (\frac{\pi }{2},\frac{3\pi }{2})$, by ③ $$\sum _{k=1}^n|z_k{|}^2=\frac{1}{2}\sum _{k=1}^n(|z_k{|}^2+|z_{k+1}{|}^2)\ge \frac{1}{4}\sum _{k=1}^n\frac{1}{{\sin }^2\frac{{\theta }_k}{2}}.④$$ Since ${\prod }_{k=1}^n\frac{z_{k+1}}{z_k}=\frac{z_{n+1}}{z_1}=1$, so $$\sum _{k=1}^n{\theta }_k=\arg \left(\prod _{k=1}^n\frac{z_{k+1}}{z_k}\right)+2m\pi =2m\pi ,⑤$$ where $m$ is a positive integer, and $m<n$. Notice that $n$ is odd, so $$0<\sin \frac{m\pi }{n}\le \sin \frac{(n-1)\pi }{2n}=\cos \frac{\pi }{2n}.⑥$$ 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 $$\begin{array}{rl}\sum _{k=1}^n|z_k{|}^2& \ge \frac{1}{4}\sum _{k=1}^n\frac{1}{{\sin }^2\frac{{\theta }_k}{2}}\\ & \ge \frac{n}{4}\cdot \frac{1}{{\sin }^2\left(\frac{1}{n}{\sum }_{k=1}^n\frac{{\theta }_k}{2}\right)}\\ & =\frac{n}{4}\cdot \frac{1}{{\sin }^2\frac{m\pi }{n}}\ge \frac{n}{4}\cdot \frac{1}{{\cos }^2\frac{\pi }{2n}}={\lambda }_0(n).\end{array}$$

(2) If there exists $j$ ($1\le j\le n$), 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 $|z_j{|}^2+|z_{j+1}{|}^2\ge 1$; and by ③, for $j\notin I$, we have $$|z_j{|}^2+|z_{j+1}{|}^2\ge \frac{1}{2{\sin }^2\frac{{\theta }_j}{2}}\ge \frac{1}{2}.$$ Therefore, $$\begin{array}{rl}\sum _{k=1}^n|z_k{|}^2& =\frac{1}{2}\left(\sum _{j\in I}(|z_j{|}^2+|z_{j+1}{|}^2)+\sum _{j\notin I}(|z_j{|}^2+|z_{j+1}{|}^2)\right)\\ & \ge \frac{1}{2}|I|+\frac{1}{4}(n-|I|)=\frac{1}{4}(n+|I|)\ge \frac{n+1}{4}.\end{array}$$

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

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 $${\lambda }_0(n)=\{\begin{array}{ll}\frac{n}{4},& 2\mid n,\\ \frac{n}{4{\cos }^2\frac{\pi }{2n}},& \text{otherwise.}\end{array}$$