July 29, 2023
OC635 (from Crux Mathematicorum Vol. 49, No. 5)
Let \(n \ge 3\) be an integer. Prove that for all positive real numbers \(x_1, \ldots ,x_n\), \[\frac{1 + x_1^2}{x_2 + x_3} + \frac{1 + x_2^2}{x_3 + x_4} + \cdots + \frac{1 + x_{n-1}^2}{x_n + x_1} + \frac{1 + x_{n}^2}{x_1 + x_2} \ge n\]
By the Cauchy–Schwarz inequality, we have \[\begin{align*} \frac{x_1^2}{x_2 + x_3} + \frac{x_2^2}{x_3 + x_4} + \cdots + \frac{x_{n-1}^2}{x_n + x_1} + \frac{x_{n}^2}{x_1 + x_2} \\ \ge \frac{(x_1 + x_2 + \ldots + x_n)^2}{2(x_1 + x_2 + \ldots + x_n)} = \frac{2(x_1 + x_2 + \ldots + x_n)}{4}. \end{align*}\] By the AM-GM inequality, we have \[\frac{1}{x_2 + x_3} + \frac{x_2 + x_3}{4} \ge 2 \sqrt{\frac{1}{x_2 + x_3} \frac{x_2 + x_3}{4}} = 1. \]
Combining the above two observations, we establish \[\begin{align*} & \frac{1 + x_1^2}{x_2 + x_3} + \frac{1 + x_2^2}{x_3 + x_4} + \cdots + \frac{1 + x_{n-1}^2}{x_n + x_1} + \frac{1 + x_{n}^2}{x_1 + x_2} \\ \ge \; & (\frac{1}{x_1 + x_2} + \frac{x_1 + x_2}{4}) + (\frac{1}{x_2 + x_3} + \frac{x_2 + x_3}{4}) + \ldots + (\frac{1}{x_n + x_1} + \frac{x_n + x_1}{4}) \\ \ge \; & n \end{align*}\] which is the desired inequality.