Let N = {q^k}{n^2} be an odd perfect number with Euler
factor q^k.
Suppose sigma(n)/q < sigma(q)/n.
Then n*sigma(n) < q*sigma(q).
Since I(q) < 5/4 < I(n) (where I(x) is the abundancy index of x),
then sigma(q)/q < sigma(n)/n. This implies that:
n*sigma(q) < q*sigma(n)
Therefore:
n*(sigma(q) + sigma(n)) < q*(sigma(q) + sigma(n))
Consequently:
n < q
Result #1: If sigma(n)/q < sigma(q)/n, then n < q.
Note that, in this case:
sigma(n)/q^k <= sigma(n)/q < sigma(q)/n <= sigma(q^k)/n.
Corollary #1: If sigma(n)/q^k < sigma(q^k)/n, then
n < q^k.
Proof of Corollary #1:
Suppose sigma(n)/q^k < sigma(q^k)/n. Then
n*sigma(n) < {q^k}*sigma(q^k)
Since I(q^k) < 5/4 < I(n), then
sigma(q^k)/q^k < sigma(n)/n, which gives:
n*sigma(q^k) < {q^k}*sigma(n)
Therefore:
n*(sigma(q^k) + sigma(n)) < {q^k}*(sigma(q^k) + sigma(n))
Consequently:
n < q^k.
Now, suppose that sigma(q)/n < sigma(n)/q. Then:
q*sigma(q) < n*sigma(n)
Again:
sigma(q)/q < sigma(n)/n, which gives:
n*sigma(q) < q*sigma(n)
Therefore:
(q + n)*sigma(q) < (q + n)*sigma(n)
Consequently:
sigma(q) < sigma(n)
Result #2: If sigma(q)/n < sigma(n)/q,
then sigma(q) < sigma(n).
Corollary #2: If sigma(q^k)/n < sigma(n)/q^k,
then sigma(q^k) < sigma(n).
Now, if sigma(q) < sigma(n) then:
q + 1 < sigma(n) < 2n
(q + 1)/2 < n