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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Algebra Difficulty 4.6 AIME Prove it Romania

Let $a_1$ , $a_2$ , $a_3$ , $a_4$ , $a_5$ be five real numbers of zero sum, such that $|a_i - a_j| \le 1$ , for all $i, j \in \{1, 2, 3, 4, 5\}$ . Prove that $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \le \frac{6}{5}$ .

Solution

Since $0 = (a_1 + a_2 + a_3 + a_4 + a_5)^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + 2 \sum_{i<j} a_i a_j$ , it follows that $\sum_{i<j} (a_i - a_j)^2 = 4(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) - 2 \sum_{i<j} a_i a_j = 5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2)$ . On the other side, $\sum_{i<j} (a_i - a_j)^2 \le \sum_{i<j} |a_i - a_j|$ , for $|a_i - a_j| \le 1$ . Assuming $a_1 \le a_2 \le a_3 \le a_4 \le a_5$ , we have $\sum_{i<j} |a_i - a_j| = 4(a_5 - a_1) + 2(a_4 - a_2) \le 4 + 2 = 6$ . So $5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) \le 6$ , as requested to be shown.

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