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Algebra Difficulty 8.9 Shortlist Prove it China

Given an integer n3n \ge 3, find the maximum real number MM, such that for any positive numbers x1,x2,,xnx_1, x_2, \dots, x_n, there exists a permutation y1,y2,,yny_1, y_2, \dots, y_n of x1,x2,,xnx_1, x_2, \dots, x_n that satisfies
i=1nyi2yi+12yi+1yi+2+yi+22M, \sum_{i=1}^{n} \frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} \ge M,
where yn+1=y1y_{n+1} = y_1, yn+2=y2y_{n+2} = y_2. (posed by Qu Zhenhua)

Solution

Let
F(x1,,xn)=i=1nxi2xi+12xi+1xi+2+xi+22. F(x_1, \dots, x_n) = \sum_{i=1}^{n} \frac{x_i^2}{x_{i+1}^2 - x_{i+1}x_{i+2} + x_{i+2}^2}.
First, take x1=x2==xn1=1x_1 = x_2 = \cdots = x_{n-1} = 1, xn=ϵx_n = \epsilon, then all permutations are the same in the sense of circulation. In this case, we have
F(x1,,xn)=n3+21ϵ+ϵ2+ϵ2. F(x_1, \dots, x_n) = n - 3 + \frac{2}{1 - \epsilon + \epsilon^2} + \epsilon^2.
Let ϵ0+\epsilon \to 0^+, Fn1F \to n-1, so Mn1M \le n-1.

Next, we show that for any positive numbers x1,,xnx_1, \dots, x_n, there exists a permutation y1,,yny_1, \dots, y_n satisfying F(y1,,yn)n1F(y_1, \dots, y_n) \ge n-1. In fact, take the permutation y1,,yny_1, \dots, y_n with y1y2yny_1 \ge y_2 \ge \dots \ge y_n and by the inequality a2ab+b2max(a2,b2)a^2 - ab + b^2 \le \max(a^2, b^2), we see that
F(y1,,yn)y12y22+y22y32++yn12y12n1, F(y_1, \dots, y_n) \ge \frac{y_1^2}{y_2^2} + \frac{y_2^2}{y_3^2} + \dots + \frac{y_{n-1}^2}{y_1^2} \ge n-1,
where the last inequality is obtained by AM-GM inequality.

Summing up, M=n1M = n-1. \square

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