Problem

Source: FKMO 2024 P3

Tags: algebra, Inequality



Find the smallest real number $p(\leq 1)$ that satisfies the following condition. (Condition) For real numbers $x_1, x_2, \dots, x_{2024}, y_1, y_2, \dots, y_{2024}$, if $0 \leq x_1 \leq x_2 \leq \dots \leq x_{2024} \leq 1$, $0 \leq y_1 \leq y_2 \leq \dots \leq y_{2024} \leq 1$, $\displaystyle \sum_{i=1}^{2024}x_i = \displaystyle \sum_{i=1}^{2024}y_i = 2024p$, then the inequality $\displaystyle \sum_{i=1}^{2024}x_i(y_{2025-i}-y_{2024-i}) \geq 1 - p$ holds.