Recall that $P(n)$ denotes the largest prime factor of a positive integer $n$. Let $A$ be the set of all positive integers such that the $P(n)$-adic valuation of $n$ is larger than $P(n)/2$. In other words, $$ A:=\{p_1^{a_1}\cdots p_k^{a_k} : p_1<\cdots <p_k, a_i\geq 0, a_k>p_k/2\}. $$
I know that this set can be written also as the disjoint union $$ A=\bigcup_{p\geq 2}\bigcup_{i\geq 1}A_{p,i}, $$ where $A_{p,i}:=\{rp^{\lfloor p/2\rfloor+i} : r\in S(p-1)\}$ (as usual, $S(m)$ denotes the set all $m$-smooth numbers). Here consider $S(0)=\emptyset$.
I would like to find a lower bound for the natural density of $A$, so I need a lower bound for $\#A(x)=\# (A\cap [1,x])$. Clearly, $$ \#A(x) = \sum_{p\geq 2}\sum_{i\geq 1}\Psi\left(\frac{x}{p^{\lfloor p/2\rfloor+i}}, p-1\right), $$ where $\Psi(x,y)=\# (S(y)\cap [1,x])$. There is known result which states that $$ \Psi(x,y)=\frac{1}{\pi(y)!}\prod_{p\leq y}\left( \frac{\log x}{\log y}\right)\left( 1+ O\left(\frac{y^2}{\log x \log y}\right) \right) $$ as $x\to \infty$.
So, it seems that something can be proved by this. However, I think I am doing very extra work and since I am not very familiarised with analytic number theory, I would like any your suggestions.