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 1 of 6: Sort and choose the floor

Previous step Next step

s=\lfloor A_n\rfloor,\qquad a_1\ge a_2\ge\cdots\ge a_n.

Detailed analysis

Reordering does not change the product. Let s be the floor of the average, and choose k so that $a_k\ge s\ge a_{k+1}$ , with the obvious endpoint conventions.