, 2011

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.