APMO

Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s = \left\lfloor A_{n} \right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.
Our inequality is equivalent to proving that
$$\begin{equation*} \frac{a_{1}!}{s!} \cdot \frac{a_{2}!}{s!} \cdot \ldots \cdot \frac{a_{k}!}{s!} \geq \frac{s!}{a_{k+1}!} \cdot \frac{s!}{a_{k+2}!} \cdot \ldots \cdot \frac{s!}{a_{n}!} . \tag{1} \end{equation*}$$
Now for $i = 1, 2, \ldots, k$, $a_{i}! / s!$ is the product of $a_{i} - s$ factors. For example, $9! / 5! = 9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A = a_{1} + a_{2} + \cdots + a_{k} - k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B = (n - k) s - (a_{k+1} + a_{k+2} + \cdots + a_{n})$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i} = n A_{n} \geq n s$, $A \geq B$. This proves the inequality.

Equality in (1) holds if and only if either:
(i) $A = B = 0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1} = a_{2} = \cdots = a_{n}$; or
(ii) $a_{1} = 1$ and $s = 0$, that is, the only factors on either side of (1) are $1$'s, which occurs if and only if $a_{i} \in \{0, 1\}$ for all $i$.

Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d = a_{n} - a_{1}$ and $m = \left| \{ i : a_{i} = a_{1} \} \right|$. Our proof is by induction on $d$.

We first do the case $d = a_{n} - a_{1} = 0$ or $1$ separately. Then $a_{1} = a_{2} = \cdots = a_{m} = a$ and $a_{m+1} = \cdots = a_{n} = a + 1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n} \right\rfloor = a$, so the inequality to be proven is just $a_{1}! a_{2}! \ldots a_{n}! \geq (a!)^{n}$, which is obvious. Equality holds if and only if either $m = n$, that is, $a_{1} = a_{2} = \cdots = a_{n} = a$; or if $a = 0$, that is, $a_{1} = \cdots = a_{m} = 0$ and $a_{m+1} = \cdots = a_{n} = 1$.

So assume that $d = a_{n} - a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence
$$a_{1} + 1, a_{2}, a_{3}, \ldots, a_{n-1}, a_{n} - 1,$$
though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n} > a_{1} + 1$,
$$\begin{aligned} a_{1}! a_{2}! \ldots a_{n}! & = (a_{1} + 1)! a_{2}! \ldots a_{n-1}! (a_{n} - 1)! \cdot \frac{a_{n}}{a_{1} + 1} \\ & \geq \left( \left\lfloor A_{n} \right\rfloor ! \right)^{n} \cdot \frac{a_{n}}{a_{1} + 1} \\ & > \left( \left\lfloor A_{n} \right\rfloor ! \right)^{n} \end{aligned}$$
which completes the proof. Equality cannot hold in this case.