Singapore Mathematical Olympiad (SMO)

We use induction on $n$. When $n=2$, it is an equality. Assume $a_1 \le a_2 \le \dots \le a_n$. Then $a_1 \le A \le a_n$. Let $A_o = (a_1 + \dots + a_{n-1})/(n-1)$. Then $A = A_o + (a_n - A_o)/n$ and $a_1 \le A_o \le A_{n-1} \le a_n$. Since $a_n - A_o \ge 0$, we use binomial expansion:
$$A^{n-1} = \left(A_o + \frac{a_n - A_o}{n}\right)^{n-1} \ge A_o^{n-1} + (n-1)A_o^{n-2} \left(\frac{a_n - A_o}{n}\right) = \frac{n-1}{n} A_o^{n-2} a_n + \frac{1}{n} A_o^{n-1}.$$
By induction hypothesis, $A_o^{n-2} \ge \frac{1}{n-1} \sum_{i=1}^{n-1} a_i \cdots \hat{a}_i \cdots a_{n-1}$.
By $AM \ge GM$, $A_o^{n-1} \ge a_1 a_2 \cdots a_{n-1}$.
Therefore, $A^{n-1} \ge \frac{1}{n} \sum_{i=1}^{n-1} a_i \cdots \hat{a}_i \cdots a_{n-1} a_n + \frac{1}{n} a_1 a_2 \cdots a_{n-1} = \frac{1}{n} \sum_{i=1}^{n} a_i \cdots \hat{a}_i \cdots a_n$.