Round 3

Algebra Difficulty 4.7 AIME Prove it Mongolia

Let $d_1, d_2, \dots, d_n$ be nonnegative real numbers satisfying $1 \ge d_1 \ge \dots \ge d_n \ge 0$ . Prove
$\frac{(1 + d_1 + d_2 + \dots + d_n)^2}{n + 1} \ge 2 \cdot \frac{d_1^2 + 2d_2^2 + \dots + n d_n^2}{n}.$

Solution

Setting $d_0 = 1$ , we have $j d_j \le d_0 + d_1 + \dots + d_{j-1}$ for any $1 \le j \le n$ . Hence
$\sum_{j=1}^{n} j d_j^2 \le \sum_{j=1}^{n} \sum_{i=0}^{j-1} d_i d_j = \sum_{0 \le i < j \le n} d_i d_j \le \frac{n}{2(n+1)} \left( \sum_{k=0}^{n} d_k \right)^2,$
where the last part follows from Maclaurin's inequality. Equality holds for $d_1 = d_2 = \dots = d_n = 1$ only.

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