China Southeastern Mathematical Olympiad

Since
$$\sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j \min(r_i, r_j) = \sum_{j=1}^{n} a_1 a_j \min(r_1, r_j) + \sum_{j=1}^{n} a_2 a_j \min(r_2, r_j) \\ + \dots + \sum_{j=1}^{n} a_k a_j \min(r_k, r_j) + \dots \\ + \sum_{j=1}^{n} a_n a_j \min(r_n, r_j),$$
its $k$-th term is
$$\sum_{j=1}^{n} a_k a_j \min(r_k, r_j) = a_k a_1 r_1 + a_k a_2 r_2 + \dots + a_k a_k r_k \\ + a_k a_{k+1} r_k + \dots + a_k a_n r_n$$
which is the sum of elements of the $k$-th row of $A_1$, $k = 1, 2, \dots, n$.
Therefore, $\sum_{i=1}^n \sum_{j=1}^n a_i a_j \min(r_i, r_j)$ is the sum of all elements of $A_1$.

On the other hand, the summation can also be done as follows: Take the elements of first column and the first row of $A_1$, sum up; denote the rest $(n-1) \times (n-1)$ element by matrix $A_2$, then take the first column and the first row of $A_2$, sum up, denote the rest by matrix $A_3, \dots$, so we get
$$\begin{align*} \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j \min(r_i, r_j) &= \sum_{k=1}^{n} r_k (a_k^2 + 2a_k (a_{k+1} + a_{k+2} + \dots + a_n)) \\ &= \sum_{k=1}^{n} r_k \left( \left(a_k + \sum_{i=k+1}^{n} a_i\right)^2 - \left(\sum_{i=k+1}^{n} a_i\right)^2 \right) \\ &= \sum_{k=1}^{n} r_k \left( \left(\sum_{i=k}^{n} a_i\right)^2 - \left(\sum_{i=k+1}^{n} a_i\right)^2 \right) \\ &= r_1 \left(\sum_{i=1}^{n} a_i\right)^2 + r_2 \left(\sum_{i=2}^{n} a_i\right)^2 + r_3 \left(\sum_{i=3}^{n} a_i\right)^2 + \dots \\ &\quad + r_n \left(\sum_{i=n}^{n} a_i\right)^2 - r_1 \left(\sum_{i=2}^{n} a_i\right)^2 - r_2 \left(\sum_{i=3}^{n} a_i\right)^2 \\ &\quad - \dots - r_{n-1} \left(\sum_{i=n}^{n} a_i\right)^2 \\ &= \sum_{k=1}^{n} (r_k - r_{k-1}) \left(\sum_{i=k}^{n} a_i\right)^2 \ge 0 \end{align*}$$
(where $r_0 = 0$).