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 4 of 6: Prove the inequality

Previous step Next step

A\ge B\implies\prod_{i=1}^{n}a_i!\ge(s!)^n.

Detailed analysis

Cancel $B$ numerator factors against the denominator: every remaining numerator factor is at least 1, proving the inequality. For equality, note separately that if $s>0$ and $A>B$ , at least one remaining factor is greater than 1; when $s=0$ , all factors can be 1, which is handled in the next step.