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Taiwan IMO Selection Camp

Algebra Difficulty 5.3 AIME, harder Prove it Taiwan

Let $x_1, \cdots, x_{100}$ be nonnegative real numbers satisfying, for $i = 1, \cdots, 100$ ,
$x_i + x_{i+1} + x_{i+2} \le 1,$
(where we denote $x_{101} = x_1, x_{102} = x_2$ ). Find the maximum possible value of
$S = \sum_{i=1}^{100} x_i x_{i+2}.$

Solution

Answer: $\frac{25}{2}$ .
Let $x_{2i} = 0, x_{2i-1} = \frac{1}{2}$ for all $i = 1, \cdots, 50$ . Then we have
$S = 50\left(\frac{1}{2}\right)^2 = \frac{25}{2}.$
Thus it suffices to show $S \le \frac{25}{2}$ for all $x_i$ satisfying the given conditions.
Consider $1 \le i \le 50$ . From the given conditions we obtain
$x_{2i-1} \le 1 - x_{2i} - x_{2i+1}, \quad x_{2i+2} \le 1 - x_{2i} - x_{2i+1}.$
Then by the AM-GM inequality we get
$\begin{aligned} x_{2i-1}x_{2i+1} + x_2x_{2i+2} &\le (1-x_{2i}-x_{2i+1})x_{2i+1} + x_2(1-x_{2i}-x_{2i+1}) \\ &= (x_{2i}+x_{2i+1})(1-x_{2i}-x_{2i+1}) \\ &\le \left(\frac{x_{2i}+x_{2i+1}}{2} + (1-x_{2i}-x_{2i+1})\right)^2 = \frac{1}{4}. \end{aligned}$
Summing the above inequality for $i = 1, \cdots, 50$ , we obtain
$\sum_{i=1}^{50} (x_{2i-1}x_{2i+1} + x_2x_{2i+2}) \le 50 \cdot \frac{1}{4} = \frac{25}{2}.$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.