China National Team Selection Test

Consider
$$D_1 = (a_2, a_3, \dots, a_n) \text{ and } d_2 = (a_2, A) = (a_2, a_1 + a_2 + \dots + a_n).$$
Let $(D_1, d_2) = d$. Then $d \mid a_2, d \mid a_3, \dots, d \mid a_n, d \mid a_1 + a_2 + \dots + a_n$. Thus, $d \mid a_1$. Consequently,
$$d \mid (a_1, a_2, \dots, a_n).$$
Since $a_1, a_2, \dots, a_n$ are coprime, we have $d = 1$. Note that $D_1 \mid a_2, D_2 \mid a_2$ and $(D_1, D_2) = 1$. We have $D_1 d_2 \mid a_2$. So $D_1 d_2 \le a_2$. Similarly, we have $D_2 d_3 \le a_3, \dots, D_n d_1 \le a_1$. Hence,
$$\begin{aligned} \prod_{i=1}^{n} d_i D_i &= (D_1 d_2) \cdot (D_2 d_3) \cdot \dots \cdot (D_n d_1) \\ &\le a_2 a_3 \dots a_n a_1 \\ &= \prod_{i=1}^{n} a_i. \end{aligned} \qquad \textcircled{1}$$
Considering
$$\begin{aligned} \prod_{i=1}^{n} (A - a_i) &= \prod_{i=1}^{n} \left( \sum_{j \neq i} a_j \right) \\ &\geq \prod_{i=1}^{n} \left( (n-1) \left( \prod_{j \neq i} a_j \right)^{\frac{1}{n-1}} \right) \\ &= (n-1)^n \cdot \prod_{i=1}^{n} a_i \end{aligned} \qquad \textcircled{2}$$
and by ① and ②, we see that
$$\prod_{i=1}^{n} \frac{A - a_i}{d_i D_i} \ge (n-1)^n.$$
On the other hand, if $a_1 = a_2 = \dots = a_n = 1$,
$$\prod_{i=1}^{n} \frac{A - a_i}{d_i D_i} = (n - 1)^n.$$
Summing up, the minimum of $\prod_{i=1}^{n} \frac{A - a_i}{d_i D_i}$ is $(n - 1)^n$.