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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.

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2 Answers 2

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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.

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Hint: You can try to find a constant $c_n$ (the optimal one) such that

$$c_n(x_1^2 + \cdots + x_n^2) \ge \sum_{\textrm{cyc}} x_i x_{i+1}$$

It has to do with the eigenvalues of the matrix that gives the quadratic form on RHS.

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  • $\begingroup$ Thanks. Would it be possible to expand on your hint? $\endgroup$
    – user33096
    Oct 8, 2022 at 13:58
  • $\begingroup$ @user33096: I wonder if you know about eigenvalues (and eigenvectors) of symmetric matrices. It is important in the theory of inequalities (Sum of squares method). $\endgroup$
    – orangeskid
    Oct 12, 2022 at 12:09

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