XIV APMO

Assume without loss of generality that $a_1\ge a_2\ge \cdots \ge a_n\ge 0$, and let $s=\lfloor A_n\rfloor$. Let $k$ be any (fixed) index for which $a_k\ge s\ge a_{k+1}$. Our inequality is equivalent to proving that $$\begin{array}{c}\frac{a_1!}{s!}\cdot \frac{a_2!}{s!}\cdot \dots \cdot \frac{a_k!}{s!}\ge \frac{s!}{a_{k+1}!}\cdot \frac{s!}{a_{k+2}!}\cdot \dots \cdot \frac{s!}{a_n!}.\end{array}\text{\hspace{1em}(1)}$$ Now for $i=1,2,\dots ,k$, $a_i!/s!$ 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 $A=a_1+a_2+\cdots +a_k-ks$ 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 ${\sum }_{i=1}^n{a}_i=n{A}_n\ge ns$, $A\ge B$. 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 $i$.

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Alternative solution.

Assume without loss of generality that $0\le a_1\le a_2\le \cdots \le a_n$. 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 $1\le m\le n$ and $a\ge 0$. In this case we have $\lfloor A_n\rfloor =a$, so the inequality to be proven is just $a_1!a_2!\dots a_n!\ge (a!{)}^n$, 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 $d=a_n-a_1\ge 2$ 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 $$a_1+1,a_2,a_3,\dots ,a_{n-1},a_n-1,$$ 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 $a_n>a_1+1$, $$\begin{array}{rl}{a}_1!a_2!\dots a_n!& =(a_1+1)!a_2!\dots a_{n-1}!(a_n-1)!\cdot \frac{a_n}{a_1+1}\\ & \ge {\left(\lfloor A_n\rfloor !\right)}^n\cdot \frac{a_n}{a_1+1}\\ & >{\left(\lfloor A_n\rfloor !\right)}^n\end{array}$$ which completes the proof. Equality cannot hold in this case.