Can this bound for the abundancy index of $n$ be improved, given that $q^k n^2$ is an odd perfect number with $k=1$? In what follows, set $I(x)=\sigma(x)/x$ to be the abundancy index of $x \in \mathbb{N}$, where $\sigma(x)$ is the sum of divisors of $x$.  If $I(y)=2$ and $y$ is odd, then $y$ is called an odd perfect number.  Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
The following result was communicated to me (via e-mail, by Pascal Ochem) on April 17, 2013.

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then
  $$I(n) > {\left(\frac{8}{5}\right)}^{\frac{\ln(4/3)}{\ln(13/9)}} \approx 1.44440557.$$

The proof of this result uses the following lemma.

Let $x(n) = \ln(I(n^2))/\ln(I(n))$.  If $\gcd(a, b) = 1$, then
  $$\min(x(a), x(b)) < x(ab) < \max(x(a),x(b)).$$

Proof of Lemma:  First, note that $x(a) \neq x(b)$ (since $\gcd(a , b) = 1$). Without loss of generality, we may assume that $x(a) < x(b)$.  Thus, since
$$x(n) = \frac{\ln(I(n^2))}{\ln(I(n))},$$
we have
$$\frac{\ln(I(a^2))}{\ln(I(a))} < \frac{\ln(I(b^2))}{\ln(I(b))}.$$
This implies that
$$\ln(I(a^2))\ln(I(b)) < \ln(I(b^2))\ln(I(a)).$$
Adding $\ln(I(a^2))\ln(I(a))$ to both sides of the last inequality, we get
$$\ln(I(a^2))\bigg(\ln(I(b)) + \ln(I(a))\bigg) < \ln(I(a))\bigg(\ln(I(b^2)) + \ln(I(a^2))\bigg).$$
Using the identity $\ln(X) + \ln(Y) = \ln(XY)$, we can rewrite the last inequality as
$$\ln(I(a^2))\ln(I(ab)) < \ln(I(a))\ln(I((ab)^2))$$
since $I(x)$ is a weakly multiplicative function of $x$ and $\gcd(a, b) = 1$.
It follows that
$$\min(x(a), x(b))  = x(a) = \frac{\ln(I(a^2))}{\ln(I(a))} < \frac{\ln(I((ab)^2))}{\ln(I(ab))} = x(ab).$$
Under the same assumption $x(a) < x(b)$, we can show that $$x(ab) < x(b) = \max(x(a),x(b))$$
by adding $\ln(I(b^2))\ln(I(b))$ ( instead of $\ln(I(a^2))\ln(I(a))$ ) to both sides of the inequality
$$\ln(I(a^2))\ln(I(b)) < \ln(I(b^2))\ln(I(a)).$$
This finishes the proof.


From this lemma, we note that $1 < x(n) < 2$ follows from
    $$I(n) < I(n^2) < {\left(I(n)\right)}^2$$
    and $I(n^2) = {\left(I(n)\right)}^{x(n)}$.
The trivial lower bound for $I(n)$ is
    $${\bigg(\frac{8}{5}\bigg)}^{1/2} < I(n).$$
Note that decreasing the denominator in the exponent gives an increase in the lower bound for $I(n)$.


Proof of Result: We sketch a proof for the mentioned result here, as communicated to the author by Pascal Ochem.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.
We want to obtain a lower bound on $I(n)$.  We know that
$$I(n^2) = 2/I(q^k) > 2/(5/4) = 8/5.$$
We need to improve the trivial bound $I(n^2) < {\left(I(n)\right)}^2$.
Let $x(n)$ be such that 
$$I(n^2) = {\bigg(I(n)\bigg)}^{x(n)}.$$
That is, $x(n) = \ln(I(n^2))/\ln(I(n))$.  We want an upper bound on $x(n)$ for $n$ odd.  By the lemma, we consider the component $r^s$ with $r$ prime that maximizes $x(r^s)$.
We have
$$I(r^s) = \frac{r^{s + 1} - 1}{{r^s}(r - 1)} = 1 + \frac{1}{r - 1} - \frac{1}{{r^s}(r - 1)}.$$
Also,
$$I(r^{2s}) = \frac{r^{2s + 1} - 1}{{r^{2s}}(r - 1)} = I(r^s)\bigg(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right)\bigg).$$
So,
$$x(r^s) = \frac{\ln(I(r^{2s}))}{\ln(I(r^s))} = \frac{\ln(I(r^s)) + \ln(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right))}{\ln(I(r^s))},$$
from which it follows that
$$x(r^s) = 1 + \frac{\ln(1 + \left(\frac{1 - r^{-s}}{r^{s + 1} - 1}\right))}{\ln(1 + \frac{1}{r - 1} - \frac{1}{{r^s}(r - 1)})}.$$
We can check that
$$x(r^s) > x(r^t)$$
if $s < t$ and $r \geq 3$.  Therefore, $x(r^s)$ is maximized for $s = 1$.
Now,
$$x(r) = 1 + \frac{\ln(1 + (1/(r(r + 1))))}{\ln(1 + (1/r))} = \frac{\ln(1 + (1/r) + (1/r)^2)}{\ln(1 + (1/r))} = \ln(I(r^2))/\ln(I(r)),$$ 
which is maximized for $r = 3$.
So,
$$x(3) = \ln(I(3^2))/\ln(I(3)) = \ln(13/9)/\ln(4/3) \approx 1.27823.$$
The claimed result then follows, and the proof is complete.
Motivation for My Inquiry
Note that, since $q$ is prime with $q \equiv 1 \pmod 4$, then $q \geq 5$.  In particular, when $k=1$, then we obtain
$$I(q^k) = 1 + \frac{1}{q} \leq \frac{6}{5} \iff I(n^2) = \frac{2}{I(q^k)} \geq \frac{5}{3},$$
from which we get
$$I(n) > \bigg(I(n^2)\bigg)^{1/{x(3)}} \geq \bigg(\frac{5}{3}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} \approx 1.4912789897463723558.$$
Here is my question:

Would it be possible to tweak the argument in this post in order to come up with a proof for the implication
  $$k=1 \implies I(n) > \frac{3}{2}?$$

 A: (Not a complete answer.)
Let $q^k n^2$ be an odd perfect number with special/Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
First off, we show the following lemma.
Lemma 1 $I(n) \neq 3/2 = I(2)$
Proof  The proof is trivial and follows from the fact that $n \neq 2$, since $n$ must be odd.  (In fact, we know by Nielsen's latest result that $\omega(n) \geq 9$, where $\omega(n)$ is the number of distinct prime factors of $n$.  In particular, this means that $n$ is composite.)  Note that $2$ is solitary as it is prime.
Next, we prove the following implication.
Theorem $\bigg((k=1) \land (I(n) < 3/2)\bigg) \implies q=5$
Proof Since $k=1$, then we obtain
$$\frac{2q}{q+1}=I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$
Using WolframAlpha, we get that $-1 < q < 5.23316$, from which we conclude that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.
In general, we can remove the assumption that $k=1$:
Theorem $I(n) < 3/2 \implies q=5$
Proof We proceed as follows:
$$\frac{2(q-1)}{q} < I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$
Using WolframAlpha, we obtain $0 < q < 6.23316$, which implies that $q=5$, since $q$ is a prime number satisfying $q \equiv 1 \pmod 4$.
Remarks: Note that $I(n) > 3/2$ would also follow from the inequality $D(n) < s(n)$, where $D(n)=2n-\sigma(n)$ is the deficiency of $n$ and $s(n)=\sigma(n) - n$ is the sum of the aliquot divisors of $n$.
