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IRL ABooklet

Algebra Difficulty 4.6 AIME Prove it Ireland

Let x1,x2,…,x2020x_1, x_2, \dots, x_{2020} be non-negative real numbers such that
xi+xi+1+xi+2≤2for i=1,2,…,2018. x_i + x_{i+1} + x_{i+2} \le 2 \quad \text{for } i = 1, 2, \dots, 2018.
Show that
∑i=12018xixi+2≤1009. \sum_{i=1}^{2018} x_i x_{i+2} \le 1009.

Solution

Let us consider the products in pairs, starting with x1x3+x2x4x_1x_3 + x_2x_4. This relates to four consecutive terms. Setting c=max⁡{x1,x4}c = \max\{x_1, x_4\} gives x1x3≤cx3x_1x_3 \le c x_3 and x2x4≤cx2x_2x_4 \le c x_2. Moreover, the assumptions imply c+x2+x3≤2c + x_2 + x_3 \le 2 and so
x1x3+x2x4≤c(x2+x3)≤(c+x2+x32)2≤1 x_1x_3 + x_2x_4 \le c(x_2 + x_3) \le \left(\frac{c + x_2 + x_3}{2}\right)^2 \le 1
using AM-GM in the middle. More generally, for i=1,2,…,1009i = 1, 2, \dots, 1009 the same logic implies x2i−1x2i+1+x2ix2i+2≤1x_{2i-1}x_{2i+1} + x_{2i}x_{2i+2} \le 1. Adding these inequalities gives the required result.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.