Proof: Let $\{t_{r}\}$ be a periodic sequence with period $n$. When $1 \leqslant k \leqslant n$, $t_{k}=b_{k}$. Let $A_{k}=t_{k+m}-b_{k}$ $(1 \leqslant k \leqslant n, m=0,1,2, \cdots, n-1)$. By (I), the sequence $\{A_{k}\}$ satisfies $S_{n}=0, S_{k} \geqslant 0$ ($1 \leqslant k \leqslant n-1$). Therefore, by Theorem (1), we have
$$T_{n}=\sum_{i=1}^{n} a_{i} A_{i}=\sum_{i=1}^{n} a_{i}\left(t_{i+m}-b_{i}\right) \leqslant 0,$$

which implies $\sum_{i=1}^{n} a_{i} b_{i} \geqslant \sum_{i=1}^{n} a_{i} t_{i+m}$, for $m=0,1,2, \cdots, n-1$. Summing both sides over $m=0,1,2, \cdots, n-1$, we get
$$n \sum_{i=1}^{n} a_{i} b_{i} \geqslant\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right).$$

This proves the right-hand side of the inequality.
$$\text{Now let } A_{i}^{\prime}=b_{n-i+1}-t_{i+m} (1 \leqslant i \leqslant n, \\ m=0,1,2, \cdots, n-1), \text{ by (II), the sequence } \{A_{i}^{\prime}\} \text{ satisfies } S_{n}=0, S_{k} \geqslant 0 (1 \leqslant k \leqslant n-1). \text{ Therefore, by Theorem (1), we have }$$
$$\begin{aligned} T_{n}^{\prime} & =\sum_{i=1}^{n} a_{i} A_{i}^{\prime}=\sum_{i=1}^{n} a_{i}\left(b_{n-i+1}-t_{i+m}\right) \\ & \leqslant 0, \end{aligned}$$
$$\begin{array}{l} \text{which implies } \sum_{i=1}^{n} a_{i} b_{n-i+1} \leqslant \sum_{i=1}^{n} a_{i} t_{i+m}, m=0, \\ 1,2, \cdots, n-1. \end{array}$$

Summing both sides over $m$ yields the left-hand side of the original inequality.