# 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.

• Hint: Show that if you have a counterexample for $n$ where one of the terms is $0$, then we can extend that counterexample to $n+1$ by adding a $0$ term. $\quad$ Hint: Play with small $n$. What's the smallest $n$ that you can find a counter example and a term of $0$? (It's almost certainly $<10$.) $\quad$ Hint: It remains to determine for small enough $n$, if the condition is ever satisfied. Mar 5, 2022 at 16:04
• What is the source of this question, please? Mar 6, 2022 at 0:36
• It's a problem from a workbook. Mar 6, 2022 at 3:53
• Thanks, but that doesn't tell me very much. A workbook for what? Does it just have problems, or also exposition? If it has exposition, what's being exposed in the part that has this problem? Mar 6, 2022 at 10:43

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 $$n\le 4.$$
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.