China National Team Selection Test

Consider $$D_1=(a_2,a_3,\dots ,a_n)\text{ and }{d}_2=(a_2,A)=(a_2,a_1+a_2+\cdots +a_n).$$ Let $(D_1,d_2)=d$. Then $d\mid a_2,d\mid a_3,\dots ,d\mid a_n,d\mid a_1+a_2+\cdots +a_n$. 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, $$\begin{array}{rl}\prod _{i=1}^n{d}_i{D}_i& =(D_1{d}_2)\cdot (D_2{d}_3)\cdot \cdots \cdot (D_n{d}_1)\\ & \le a_2{a}_3\dots a_n{a}_1\\ & =\prod _{i=1}^n{a}_i.\end{array}\stackrel{◯}{1}$$ Considering $$\begin{array}{rl}\prod _{i=1}^n(A-a_i)& =\prod _{i=1}^n\left(\sum _{j\ne i}{a}_j\right)\\ & \ge \prod _{i=1}^n\left((n-1){\left(\prod _{j\ne i}{a}_j\right)}^{\frac{1}{n-1}}\right)\\ & =(n-1{)}^n\cdot \prod _{i=1}^n{a}_i\end{array}\stackrel{◯}{2}$$ 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 $a_1=a_2=\cdots =a_n=1$, $$\prod _{i=1}^n\frac{A-a_i}{d_i{D}_i}=(n-1{)}^n.$$ Summing up, the minimum of ${\prod }_{i=1}^n\frac{A-a_i}{d_i{D}_i}$ is $(n-1{)}^n$.