Let $N = {q^k}{n^2}$ be an odd perfect number (hereinafter abbreviated as OPN) given in Eulerian form (i.e. $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).
From previous posts, we know that $n < q \Longrightarrow k = 1$.
We also know that:
$$I(q) \leq 6/5 < \sqrt{5/3} < I(n)$$
In particular, we know that:
$$\sigma(q)/\sigma(n) < q/n$$
Suppose that $\sigma(n) \leq \sigma(q)$.
It follows that:
$n < q$
and that:
$\sigma(n)/q < \sigma(q)/n$
However, from previous posts, we also know that:
$\sigma(n)/q < \sigma(q)/n \Longrightarrow n < q$
Hence, we have proved the following lemma:
Lemma 1: Let $N = {q^k}{n^2}$ be an OPN given in Eulerian form. If $\sigma(n) \leq \sigma(q)$, then $n < q$ if and only if $\sigma(n)/q < \sigma(q)/n$.
The contrapositive of:
$\sigma(n) \leq \sigma(q) \Longrightarrow n < q$
is:
$q < n \Longrightarrow \sigma(q) < \sigma(n)$
Therefore, if $q < n$, it follows that:
$\sigma(q)/n < \sigma(n)/q$
However, from previous posts, we also know that:
$\sigma(q)/n < \sigma(n)/q \Longrightarrow \sigma(q) < \sigma(n)$
Consequently, we have proved the following lemma:
Lemma 2: Let $N = {q^k}{n^2}$ be an OPN given in Eulerian form. If $q < n$, then $\sigma(q) < \sigma(n)$ if and only if $\sigma(q)/n < \sigma(n)/q$.
Note that, if $\sigma(q) < \sigma(n)$, then the best inequality that we could get relating $q$ and $n$ is:
$(q + 1)/2 < n$
This is because the upper bound $I(n) < 2$ is rather crude.