Show inequality for the number of different prime factors We consider the function $k(n) $ that represents the number of the different prime factors of $n$.We want to show that for $n>2$ 
$$k(n) \leq \frac{\log n}{\log \log n}(1+O((\log \log n)^{-1})) .$$
It seems really difficult.
Any ideas?
 A: Hint: The $k$th prime number is at least $k$, so if a number has $k$ different prime factors then it is at least $k!$. Using Stirling's approximation, you can show that $k! \leq n$ implies $k \lesssim \log n/\log\log n$, basically since $k! \approx (k/e)^k$.
A: Let $m=2\cdot 3\cdot 5\cdot\ldots\cdot p_n$. 
Then $m$ is the number with the greatest number of prime factors in the interval $[2,m]$ and:
$$\log m =\sum_{j=1}^{n}\log p_j \geq \sum_{j=1}^{n} \log((1-\varepsilon) j \log j)= n\log n+O(n\log\log n)$$
$$\log m =\sum_{j=1}^{n}\log p_j \leq \sum_{j=1}^{n} \log((1+\varepsilon) j \log j)= n\log n+O(n\log\log n)$$
by the Stirling approximation and the PNT in the form $p_n\sim n\log n$. So we have:
$$\log \log m = \log n + \log\log n + O\left(\frac{\log\log n}{\log n}\right)$$
and:
$$ \frac{\log m}{\log \log m}=n\left(1+O\left(\frac{\log\log n}{\log n}\right)\right)$$
so:
$$ n =\frac{\log m}{\log \log m}\left(1+O\left(\frac{\log\log n}{\log n}\right)\right)=\frac{\log m}{\log \log m}\left(1+O\left(\frac{\log\log\log m}{\log\log m}\right)\right).$$
It looks like replacing $\log\log\log m$ with $1$ still requires a little effort.
