Let $ n $ denote a square free positive composite integer, $ \omega(n) $ its number of prime factors, $ P\_{i}(n) $ its $ i $ -th prime factor.


Can we determine an asymptotics for the number of $ n $ below $ x $ such that $ P\_{\omega(n)}-P\_{1}(n)<\left(\dfrac{n}{\omega(n)}\right)^{1/\omega(n)} $? 



Asymptotics for a peculiar kind of squarefree numbers
Please ignore the previous version of this answer.


Motivated by Lucia's comment, we use smooth numbers to show that the number in question is $o(x)$. First note that for $100\%$ of all integers one has 
$P\_1(n)\leq \log \log \log n$, as a typical application of the naive Eratosthenes' sieve. Also by Hardy-Ramanujan one has for $100\%$ of all integers $n\leq x$ that $$ n^{1/\omega(n)}\leq x^{2/\log \log x}.$$ Therefore $$ p|n,n\leq x\Rightarrow p\leq x^{2/\log \log x}+\log \log \log x \leq x^{3/\log \log x}.$$ Now by estimates $(1,7),(1.8)$ here <http://www.dms.umontreal.ca/~andrew/PDF/msrire.pdf>
one has that for every $A>0$ the number of integers $n\leq x$ in the question is at most $$\ll x \left(\frac{3\mathrm{e}}{(\log \log x)(\log \log \log x)}\right)^{(\log \log x)/3}\ll\_A \frac{x}{(\log x)^A}.$$


