Let $\sigma_1(n)$ be the sum of the divisors of $n \in \mathbb{N}$, and let
$$I(n) = \frac{\sigma_1(n)}{n}$$
be the abundancy index of $n$.
What proportion of the natural numbers satisfy the following inequalities?
$$\frac{4n^2 + n - 1}{2n^2 + 2n} \leq I(n) < \frac{4n^2 + 2n - 1}{2n^2 + 2n}$$
In general, what approach is (most) feasible in answering these types of questions?
The problem might be ambitious, because for example it is known that certain values of $I(n)\le 2$ are connected with the existence of odd perfect numers.
Here are just some obvious remarks: We have $I(n)<2$ for all $n\ge 1$ by the inequality of the right hand side. The lower bound also tends to $2$, so that we are looking for integers $n$ with $I(n)=2-\epsilon$.
Clearly no prime number except for $p=2$ can satisfy the inequalities, because $I(p)=(p+1)/p$ is too small otherwise. Also, for all $n$ divisible by $6$ we have $I(n)\ge 2$, which is impossible. Hence all $n$ with $6\mid n$ are not possible. More generally, if $m\mid n$ then $I(n)\ge I(m)$, so that no integer $n$ can satisfy the inequalities which is divisible by $m$ with $I(m)\ge 2$. On the other hand, all powers of $2$ satisfy the inequalities, because $\sigma(2^k)=2^{k+1}-1.$