MathLabs

Problem 1

Let a1,a2,…,ana_1,a_2,\dots,a_n be a sequence of non-negative integers, and let An=(a1+a2+⋯+an)/nA_n=(a_1+a_2+\cdots+a_n)/n. Prove that a1!a2!⋯an!≥(⌊An⌋!)na_1!a_2!\cdots a_n!\ge (\lfloor A_n\rfloor!)^n, where a!=1⋅2⋯aa!=1\cdot2\cdots a for a≥1a\ge1 and 0!=10!=1. When does equality hold?
Step 2 of 6: Compare the excess counts
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A=∑i=1k(ai−s),B=∑i=k+1n(s−ai),A≥B.A=\sum_{i=1}^{k}(a_i-s),\qquad B=\sum_{i=k+1}^{n}(s-a_i),\qquad A\ge B.
Detailed analysis

Because the average is at least s, the total excess above s is at least the total deficit below s.