Prove that
$$\begin{array}{c} s_{i}=b_{1}+b_{2}+\cdots+b_{i} \\ s_{i}^{\prime}=b_{j_{1}}+b_{j_{2}}+\cdots+b_{j_{i}}(i=1,2, \cdots, n) \\ s_{i} \leqslant s_{i}^{\prime}(i=1,2, \cdots, n-1) \\ s_{n}=s_{n}^{\prime} \end{array}$$

From the given conditions, it is easy to see

Also, since $a_{i}-a_{i+1} \leqslant 0$, it follows that $s_{i}\left(a_{i}-a_{i+1}\right) \geqslant s_{i}^{\prime}\left(a_{i}-a_{i+1}\right)$.

Therefore,
$$\begin{aligned} \sum_{i=1}^{n} a_{i} b_{i} & =\sum_{i=1}^{n-1} s_{i}\left(a_{i}-a_{i+1}\right)+a_{n} s_{n} \\ & \geqslant \sum_{i=1}^{n-1} s_{i}^{\prime}\left(a_{i}-a_{i+1}\right)+a_{n} s_{n}^{\prime} \\ & =\sum_{i=1}^{n} a_{i} b_{j_{i}} \end{aligned}$$

This is the left-hand inequality. Similarly, the right-hand inequality can be proven.