TITLE: How prove this number theory inequality $\left(\dfrac{1}{N}\sum_{n=1}^{N}(\omega{(n)})^k\right)^{\frac{1}{k}}\le k+\sum_{q\le N}\frac{1}{q}$
QUESTION [6 upvotes]: show that: for any positive numbers $k$ and $N$, have
$$\left(\dfrac{1}{N}\sum_{n=1}^{N}(\omega{(n)})^k\right)^{\frac{1}{k}}\le k+\sum_{q\le N}\dfrac{1}{q}$$
where $\displaystyle\sum_{q\le N}$ is meaning no more than $N$  prime power q summation(including
$q=1$),and Let $\omega{(n)}$  denote the number of distinct prime factors of a positive integer $n$
maybe this problem background is K-th mean value Estimate of Number of prime divisors of integer see
http://www.doc88.com/p-703867145586.html
Thank you,This  is 2014 china TST test problem , maybe is old reslut? 
Thank you

REPLY [2 votes]: The solution is more or less already written on your notes.
Since $ \omega(n) = \sum_{p|n} 1$, you have:
$$\omega(n)^k = \left(\sum_{p|n} 1\right)^k = \\= k\cdot\!\!\!\!\!\!\!\!\!\sum_{\substack{p_1,\ldots,p_k|n\\ p_1\neq p_2\neq\ldots\neq p_k}}\!\!\!\!\!\!\!\!1 + \binom{k}{2}\cdot\!\!\!\!\!\!\!\!\sum_{\substack{p_1,\ldots,p_k|n\\ p_1\neq p_2\neq\ldots\neq p_k}}\!\!\!\!\!\!\!\!1 + \ldots + \binom{k}{k-1}\cdot\!\!\!\sum_{\substack{p_1,p_2|n\\ p_1\neq p_2}}\!\!1+\sum_{p|n}1.$$
Now summing each term over $n\neq N$ and using the quite crude bound:
$$\sum_{\substack{p_1\cdot\ldots\cdot p_j\leq N\\ p_1\neq p_2\neq\ldots\neq p_j}}\!\!\!\frac{1}{p_1\cdot\ldots\cdot p_j}\leq\left(\sum_{p\leq N}\frac{1}{p}\right)^j=(\log\log N+\gamma)^j$$
you get:
$$\frac{1}{N}\sum_{n\leq N}\omega(n)^k\leq (\log\log N + 1+\gamma)^k.$$