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Number of prime factors vs prime powers

Source: 2014 China TST 2 Day 1 Q1

March 20, 2014
floor functioninequalitiesnumber theory proposednumber theory

Problem Statement

Prove that for any positive integers kk and NN, (1N∑n=1N(ω(n))k)1k≤k+∑q≤N1q,\left(\frac{1}{N}\sum\limits_{n=1}^{N}(\omega (n))^k\right)^{\frac{1}{k}}\leq k+\sum\limits_{q\leq N}\frac{1}{q}, where ∑q≤N1q\sum\limits_{q\leq N}\frac{1}{q} is the summation over of prime powers q≤Nq\leq N (including q=1q=1). Note: For integer n>1n>1, ω(n)\omega (n) denotes number of distinct prime factors of nn, and ω(1)=0\omega (1)=0.
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