Let $\sigma(x)$ denote the classical sum-of-divisors function, and let

$$I(x) = \frac{\sigma(x)}{x}$$

be the *abundancy index* of the positive integer $x$.

My question is this: What proportion of the positive integers satisfy

$$I(n^2) < (1 + \frac{1}{n})I(n),$$

if, in addition, we know that both $n$ and $n^2$ are deficient numbers?

Note that, trivially, we have $I(n) \leq I(n^2) \leq (I(n))^2$ for all integers $n \geq 1$.

[This question was cross-posted from MSE.]

Thanks!