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 6 of 6: Second equality case

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s=0,\quad a_i\in\{0,1\}\ (1\le i\le n).

Detailed analysis

The other possibility is s=0 and every numerator factor is 1, which is exactly that every entry is 0 or 1. These sequences also give equality because every factorial is 1.