Let $\mathbb{N}$ be the set of positive integers.
For $x \in \mathbb{N}$, $\sigma_1(x)$ gives the sum of the divisors of $x$. (For example, $\sigma_1(3) = 1 + 3 = 4$.)
We call the ratio $I(x) = \sigma_1(x)/x$ as the abundancy index of $x$.
Since $1 \leq x \mid x^2$, by a property of the abundancy index, we know that $I(x) \leq I(x^2)$, where equality holds if and only if $x = 1$.
My question is this: Is it possible to improve on the (equivalent) inequality $x\sigma_1(x) \leq \sigma_1(x^2)$ for $x \in \mathbb{N}$?
I am guessing it would suffice to consider composite $x$.
Update: Per Will Jagy's answer (and a subsequent comment by Erick Wong), we have the conjectured (sharp?) bounds
$$1 \leq \frac{I(x^2)}{I(x)} \leq \prod_{p}{\frac{p^2 + p + 1}{p^2 + p}} = \frac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots$$
I would still be interested in an (improved) lower bound for $I(x^2)/I(x)$ when $\omega(x) \geq 3$, where $\omega(x)$ is the number of distinct prime factors of $x$.
