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Clock-tower School Seniors Competition, 2011

Algebra Difficulty 5.8 AIME, harder Prove it Romania

Let $n$ be a positive integer and $a_1 \le a_2 \le \cdots \le a_n$ be positive real numbers. Show that
$\left( \sum_{k=1}^{n} a_k^2 \right) \left( \sum_{k=1}^{n} k a_k \right) \le \left( \sum_{k=1}^{n} a_k \right) \left( \sum_{k=1}^{n} k a_k^2 \right).$

Solution

The hypothesis leads to $(i-j)(a_i - a_j)a_i a_j \ge 0$ for every $i$ and $j$ , hence
$\begin{align*} 0 &\le \sum_{i,j=1}^{n} (i-j)(a_i - a_j)a_i a_j \\ &= \sum_{i,j=1}^{n} (i a_i^2 a_j - i a_i a_j^2 - j a_i^2 a_j + j a_i a_j^2) \\ &= \left(\sum_{i=1}^{n} i a_i^2\right) \sum_{j=1}^{n} a_j - \left(\sum_{i=1}^{n} i a_i\right) \sum_{j=1}^{n} a_j^2 - \left(\sum_{i=1}^{n} a_i^2\right) \sum_{j=1}^{n} j a_j + \left(\sum_{i=1}^{n} a_i\right) \sum_{j=1}^{n} j a_j^2 \\ &= 2 \left(\sum_{k=1}^{n} k a_k^2\right) \sum_{k=1}^{n} a_k - 2 \left(\sum_{k=1}^{n} k a_k\right) \sum_{k=1}^{n} a_k^2, \end{align*}$
which leads to the required relation.

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