China Southeastern Mathematical Olympiad

Since $$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n{a}_i{a}_j\min (r_i,r_j)=\sum _{j=1}^n{a}_1{a}_j\min (r_1,r_j)+\sum _{j=1}^n{a}_2{a}_j\min (r_2,r_j) \\ +\cdots +\sum _{j=1}^n{a}_k{a}_j\min (r_k,r_j)+\dots \\ +\sum _{j=1}^n{a}_n{a}_j\min (r_n,r_j), \end{gathered}$$ its $k$-th term is $$\begin{gathered} \sum _{j=1}^n{a}_k{a}_j\min (r_k,r_j)=a_k{a}_1{r}_1+a_k{a}_2{r}_2+\cdots +a_k{a}_k{r}_k \\ +a_k{a}_{k+1}{r}_k+\cdots +a_k{a}_n{r}_n \end{gathered}$$ 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 $$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^n{a}_i{a}_j\min (r_i,r_j)& =\sum _{k=1}^n{r}_k(a_k^2+2a_k(a_{k+1}+a_{k+2}+\cdots +a_n))\\ & =\sum _{k=1}^n{r}_k\left({\left(a_k+\sum _{i=k+1}^n{a}_i\right)}^2-{\left(\sum _{i=k+1}^n{a}_i\right)}^2\right)\\ & =\sum _{k=1}^n{r}_k\left({\left(\sum _{i=k}^n{a}_i\right)}^2-{\left(\sum _{i=k+1}^n{a}_i\right)}^2\right)\\ & =r_1{\left(\sum _{i=1}^n{a}_i\right)}^2+r_2{\left(\sum _{i=2}^n{a}_i\right)}^2+r_3{\left(\sum _{i=3}^n{a}_i\right)}^2+\dots \\ & +r_n{\left(\sum _{i=n}^n{a}_i\right)}^2-r_1{\left(\sum _{i=2}^n{a}_i\right)}^2-r_2{\left(\sum _{i=3}^n{a}_i\right)}^2\\ & -\cdots -r_{n-1}{\left(\sum _{i=n}^n{a}_i\right)}^2\\ & =\sum _{k=1}^n(r_k-r_{k-1}){\left(\sum _{i=k}^n{a}_i\right)}^2\ge 0\end{array}$$ (where $r_0=0$).