Can we prove this inequality in another way?

Question

As explained here, I've managed to prove the following inequality: $\sigma(n)\geq\sqrt n(d(n)-2)+n+1$.
This can be proved easily in two cases (one for $n$ being a perfect square and one for otherwise) by summing up $n$ and $1$ and taking the AM-GM between two of each $d(n)-2$ remaining divisors of $n$, and the equality holds infinitely often, only when $n$ is a prime or a square of a prime. However, this inequality can be rewritten as following:

$\sigma(n)\geq\sqrt n(d(n)-2)+n+1$ $\Rightarrow$ $\sigma(n)\geq\sqrt n.d(n)+n+1-2\sqrt n$ $\Rightarrow$ $\sigma(n)\geq\sqrt n.d(n)+(\sqrt n-1)^2$

Which is a little cleaner but isn't obvious in the proof. Is there any other proof for this inequality so that it will lead directly to the last inequality, without the algebraic steps you see above?

I would appreciate any help :)