Problem 1

Let

a_1,a_2,\dots,a_n

be a sequence of non-negative integers, and let

A_n=(a_1+a_2+\cdots+a_n)/n

. Prove that

a_1!a_2!\cdots a_n!\ge (\lfloor A_n\rfloor!)^n

, where

a!=1\cdot2\cdots a

for

a\ge1

and

0!=1

. When does equality hold?

Step 3 of 6: Rewrite around s

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\frac{\prod_{i=1}^{n}a_i!}{(s!)^n}=\frac{\prod_{i=1}^{k}\prod_{j=1}^{a_i-s}(s+j)}{\prod_{i=k+1}^{n}\prod_{j=0}^{s-a_i-1}(s-j)}.

Detailed analysis

Factor every factorial into s! times the factors above or below s. The numerator has A factors, each at least s+1; the denominator has B factors, each at most s.