Let's assume $w(n)=\sum_{p|n}1$ - number of distinct prime divisors of $n$
I need to prove that when $n>3$
$w(n)=\mathcal{O}(\frac{lnn}{ln(lnn)})$
I know that's a classical result, however I didn't find a proof of it and I couldn't prove it myself neither. I want to find an elementary prove which doesn't use Prime Number Theorem.
Using hints I could prove such simple facts that: $w(n) \leq lnn$; $w(n)\leq\frac{2lnn}{ln(w(n))}$
But I can't combine them together into a complete proof of the fact that I need.
Among the numbers with exactly $N$ prime factors, the primorial $P_N=2\cdot 3\cdot\ldots\cdot p_N$ is the smallest one.
The problem of estimating $\omega$ turns into the problem of estimating the magnitude of $P_N$:
$$ \log P_N = \sum_{k=1}^{N} \log p_k. $$ Using Chebyshev's trick, $\binom{2M}{M}$ is a multiple of each prime in the interval $(M,2M]$, hence $$ \log P_N \leq \sum_{h=1}^{\lfloor \log_2 p_N\rfloor}\log\binom{2^{h+1}}{2^h}\leq\sum_{h=1}^{\lfloor \log_2 p_N\rfloor}2^h\log4\leq 4 p_N.$$ Chebyshev's argument gives $\log(2)\frac{n}{\log n}\leq\pi(n)\leq \log(4)\frac{n}{\log n}$, hence $$ \frac{p_N}{\log p_N} \leq \frac{N}{\log 2},\quad p_N\ll N\log N $$ and $\log P_N\ll N\log N$. It follows that in order to have $\omega(M)\geq N$, $M$ must be larger than $\exp\left(C N\log N\right)$. $$ \omega(M) \ll \frac{\log M}{\log\log M} $$ is a direct consequence.