# find all $n\ge 2$ so that $(x_1+x_2+\cdots + x_n)^2 \ge n(x_1x_2+x_2x_3+\cdots + x_nx_1)\,\forall x_1,\cdots, x_n\in \mathbb{R}_{\ge 0}$.

Determine all the values of the positive integer $$n\ge 2$$ so that $$(x_1+x_2+\cdots + x_n)^2 \ge n(x_1x_2+x_2x_3+\cdots + x_nx_1)$$ for all nonnegative real numbers $$x_1,\cdots, x_n$$.

Denote the LHS by $$F(x_1,\cdots, x_n)$$ and the RHS by $$G(x_1,\cdots, x_n)$$.

Obviously $$n=2$$ works since $$(x_1+x_2)^2 \ge 4x_1x_2 = G(x_1,x_2)$$.

Note that the expression $$(x_1+\cdots + x_n)^2 - n(x_1x_2+\cdots + x_nx_1)$$ is convex in each $$x_i$$. For convenience, let $$x_{i+n} = x_i$$ for any integer i, so on any bounded closed interval it would achieve its maximum or minimum at a boundary point. As a specific example, take $$n = 3.$$ Then we need to satisfy $$(x_1+x_2+x_3)^2 \ge 3(x_1x_2+x_2x_3+x_3x_1)$$ for all nonnegative real numbers $$x_1,\cdots, x_n$$. This holds by adding $$2(x_1x_2+x_1x_3+x_2x_3)$$ to both sides of $$x_1^2 + x_2^2+x_3^2 \ge x_1x_2+x_1x_3+x_2x_3$$. For $$n=4,$$ we must satisfy that for all real numbers $$x_1,\cdots, x_n, (x_1+x_2+x_3+x_4)^2 \ge 4(x_1x_2+\cdots + x_4x_1)$$.

I think one can induct on $$m$$ to show that for all nonnegative real numbers $$x_1,\cdots, x_m, (x_1+\cdots + x_m)^2 \ge 4(x_1x_2+\cdots + x_{m-1}x_m +x_mx_1).$$ It holds for $$m=1$$ and $$m=2$$ so suppose it holds for all $$m < k,$$ some $$k\ge 3$$. Then by cyclically shifting we may assume WLOG that $$x_k\le x_1,\cdots, x_{k-1}$$. If we have $$(x_1+\cdots + x_k)^2 = (x_1+\cdots + x_{k-1})^2 + 2x_k (x_1+\cdots + x_{k-1})+x_k^2 \ge 4(x_1x_2+\cdots + x_{k-1}x_1) + 4x_1(x_k-x_{k-1})+4x_{k-1}x_k,$$ we're done but I'm not sure how to prove this. I'm not sure about the general case for $$n\ge 5$$ though, though I think that as n gets larger the inequality will be less likely to hold. It might be useful to convert the inequality to a sum of squares somehow.

Let us first consider the case $$x_1=x_2=1$$ and $$\forall k>2, x_k=0$$. Then we need $$4 \geqslant n$$, so clearly for $$n>4$$ this doesn't hold true in general.
You have already proven it works for $$n=2, 3$$. Further, by AM-GM, note $$\frac{\color{red}{x_1+x_3}+\color{blue}{x_2+x_4}}2 \geqslant \sqrt{(\color{red}{x_1+x_3})(\color{blue}{x_2+x_4})}$$ $$\implies (x_1+x_2+x_3+x_4)^2\geqslant4(x_1x_2+x_2x_3+x_3x_4+x_4x_1)$$ Hence this is true for $$n=4$$ as well.
Hint: You can try to find a constant $$c_n$$ (the optimal one) such that
$$c_n(x_1^2 + \cdots + x_n^2) \ge \sum_{\textrm{cyc}} x_i x_{i+1}$$