On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number MOTIVATION
Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2,$$
i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what follows, I let
$$I(x)=\frac{\sigma(x)}{x}$$
denote the abundancy index of the positive integer $x$.  ($\sigma(x)$ is the sum of divisors of $x$.)
CLAIM
If $q^k n^2$ is an odd perfect number given in Eulerian form, then
$$I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5).$$
PROOF OF CLAIM
Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Suppose that
$$I(n^2) = 2 - \frac{5}{3q}.$$
Since $q^k n^2$ is perfect, then we have
$$I(q^k n^2) = I(q^k)I(n^2) = 2$$
where we have used the fact that $I$ is multiplicative.
Hence,
$$I(n^2) = I(q^k)I(n^2) - \frac{5}{3q} \implies \frac{5}{3q} = I(n^2)\bigg(I(q^k) - 1\bigg) \geq I(n^2)\bigg(1+\frac{1}{q}-1\bigg).$$
This implies that
$$I(n^2) \leq \frac{5}{3}.$$
Assume to the contrary that
$$I(n^2) = 2 - \frac{5}{3q} < \frac{5}{3}.$$
Then we have
$$\frac{6q - 5}{3q} < \frac{5}{3} \implies 18q - 15 < 15q \implies 3q < 15 \implies q < 5,$$
contradicting $q \geq 5$.
Added to the Proof of Claim (Dec 15 2019) Hence,
$$I(n^2) = 2 - \frac{5}{3q} \implies I(n^2) = \frac{5}{3} \implies (k=1 \land q=5)$$
while the proof of the direction
$$(k=1 \land q=5) \implies I(n^2) = \frac{5}{3} \implies I(n^2) = 2 - \frac{5}{3q}$$
is trivial.
QUESTION
It can be proved (page 17) that
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general for an odd perfect number $q^k n^2$ given in Eulerian form.

My question is:  Can a biconditional similar to the one proved here be derived for the case
$$I(n^2) < 2 - \frac{5}{3q}?$$

 A: Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general, and since the biconditional
$$\bigg(I(n^2) = 2 - \frac{5}{3q}\bigg) \iff \bigg(k=1 \land q=5\bigg)$$
holds, the simplest biconditional that I could come up with is
$$\bigg(I(n^2) < 2 - \frac{5}{3q}\bigg) \iff \bigg(k>1 \lor q>5\bigg).$$
Of course, by Material Implication, we also have the biconditionals
$$\bigg(k>1 \lor q>5\bigg) \iff \bigg(k=1 \implies q>5\bigg) \iff \bigg(q=5 \implies k>1\bigg).$$
A: Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since the equation
$$I(n^2) = 2 - \frac{5}{3q}$$
is true if and only if the conjunction
$$k=1 \land q=5$$
holds, then $I(n^2) = 2 - {5/(3q)}$ is true if and only if
$$I(n^2) = 2 - \frac{5}{3\cdot{5}} = 2 - \frac{1}{3} = \frac{5}{3}.$$
It follows that
$$I(n^2) < 2 - \frac{5}{3q} \iff I(n^2) \neq \frac{5}{3}.$$
Suppose that $I(n^2) > 5/3$.  We get
$$2 - \frac{5}{3q} > I(n^2) > \frac{5}{3} \implies q > 5.$$
(Another way to get the same conclusion is to use the inequality
$$I(n^2) \leq \frac{2q}{q+1},$$
which holds generally.)
Now, assume that $I(n^2) < 5/3$.  We obtain
$$\frac{2(q-1)}{q} < I(n^2) < \frac{5}{3} \implies q < 6 \implies q = 5.$$
Thus,
$$I(5^k) = I(q^k) = \frac{2}{I(n^2)} > \frac{6}{5} \implies k > 1.$$
We summarize our results below:

If $I(n^2) > 5/3$, then $q > 5$.
If $I(n^2) < 5/3$, then $q = 5$ and $k > 1$.

Since Cohen and Sorli (2012, page 4) have proved that $q=5$ and $k=5$ is untenable, then we have
$$I(n^2) < \frac{5}{3} \implies \bigg(q=5 \land k>1\bigg) \implies \bigg(q=5 \land k \geq 9\bigg) \implies I(n^2) < \frac{2}{I(5^9)} = \frac{1953125}{1220703} \approx 1.60000016384,$$
whence there is no contradiction, since it is known (unconditionally) that $I(n^2) > 8/5$ holds.
