Find $n\ge 3$ such that $\displaystyle\sum_{k=1}^n x_k = 0$ implies $\displaystyle\sum_{\mathrm{cyc}} x_1x_2\le 0$ 
Find the number of positive integers $n \ge 3$ that have the following property: If $x_1,$ $x_2,$ $\dots,$ $x_n$ are real numbers such that $x_1 + x_2 + \dots + x_n = 0,$ then
$$x_1 x_2 + x_2 x_3 + \dots + x_{n - 1} x_n + x_n x_1 \le 0.$$


I
tried using Cauchy-Schwarz to get $$\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(x_2^2+x_3^2+\cdots+x_n^2+x_1^2\right)\geq(x_1 x_2 + x_2 x_3 + \dots + x_{n - 1} x_n + x_n x_1)^2,$$ but I don't think this helps, and I'm not sure how to proceed.
 A: Consider the operator $A$ acting on $\mathbb{R}^n$ according to the rule $$(Ax)_k=x_{k-1}+x_{k+1}\quad 1\le k \le n,$$ with the convention $x_{0}=x_n$ and $x_{n+1}=x_1.$ Observe that $$ \langle Ax,x\rangle =2(x_1x_2+x_2x_3+\ldots +x_{n-1}x_n+x_nx_1).$$ Let $\theta ={2m\pi\over n},$ where $0\le m\le n-1.$ Consider the sequence  $x^\theta$ with terms $x^\theta_k=\cos k\theta.$ As $x^\theta_0=1=x^\theta_n$ and $x^\theta_{n+1}=x^\theta_1$ in a natural way, we get $$Ax^\theta=(2\cos \theta )\,x^\theta,\quad \langle Ax^\theta,x^\theta\rangle = 2\cos \theta\, \|x^\theta\|^2.$$ The matrix $A$ is symmetric, therefore the vectors $x^\theta$ are orthogonal to each other for different values of $\cos\theta.$ For $\theta=0$ we get $x^0=(1,1,\ldots, 1).$ The question arises whether  there is $\theta\neq 0, $ such that $0<\cos\theta<1.$ For $n\le 4$ it does not occur. But for $n=5$ it occurs with $\theta ={2\pi\over 5}.$ It also occurs for any $n\ge 5$ with $\theta ={2\pi \over n}.$    Summarizing the desired inequality is true in general only for $n\le 4.$
A: When $n = 3, 4$, true.
When $n \ge 5$, letting $x_{n - 1} = 0$ and $x_1 = 0$, we have:
$$x_2 + x_3 + \cdots + x_{n - 3} + x_{n - 2} = - x_n$$
implies
$$x_2x_3 + x_3 x_4 + \cdots + x_{n - 3}x_{n - 2} \le 0.$$
Not true clearly.
