Is it true that $\sigma(x^2) \geq 1 + x^2 + (\sigma(x) - 1)$ for $x > 1$?

Question

Problem

Is it true that $\sigma(x^2) \geq 1 + x^2 + (\sigma(x) - 1)$ for $x > 1$?

Here, $\sigma(N)$ is the sum of the divisors of $N \in \mathbb{N}$.

Proof Attempt

Since $1 \mid x^2$ and $x^2 \mid x^2$, then $1 + x^2$ composes part of the sum $$\sigma(x^2) = \sum_{d \mid x^2}{d}.$$

Now, because the set $D_1 = \left\{d \mid d \text{ divides } x\right\}$ is a subset of $D_2 = \left\{d \mid d \text{ divides } x^2\right\}$, then $$\sigma(x^2) \geq 1 + x^2 + (\sigma(x) - 1),$$ since $1 \in D_1 \cap D_2$.

QED

Question

Is this proof correct?

Answer 1

I am quite sure that this proof is correct but you should explain better why you have to subtract $1$ from $\sigma (x)$.

This is not because $1\in D_1$ and $1\in D_2$ but rather because you added the divisor $1$ to your sum in the first step already (when you mentioned, that $1\, |\, x^2$ and $x^2\,|\, x^2$). But $1$ is also counted in $\sigma (x)$ which is the reason why you have to subtract it. Otherwise you would have counted it twice.

Apart from that, I think the proof is totally valid.