The topic of odd perfect numbers likely needs no introduction.
Euler proved that a hypothetical odd perfect number $N$, if one exists, must have the so-called Eulerian form $N=q^k n^2$, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is known that $$q < \frac{2n^2}{D(n^2)} \leq q + 1,$$ where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. (Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.)
It has been recently shown that $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}.$$ This is equivalent to $$q\bigg(2 - I(n^2)\bigg) < 2 - \frac{1}{n^2}.$$
Since $$\frac{1}{2 - I(n^2)} \leq \frac{q+1}{2},$$ then the assumption $$\frac{q+1}{2} \leq \frac{q}{2 - (1/n^2)}$$ is equivalent to $k \neq 1$, since we then have $$\frac{q+1}{2} \leq \frac{q}{2 - (1/n^2)} < \frac{1}{2 - I(n^2)} = \frac{n^2}{D(n^2)} \leq \frac{q+1}{2},$$ whence the equation $$\frac{n^2}{D(n^2)} = \frac{q+1}{2}$$ cannot be satisfied. This shows the equivalence between $k \neq 1$ and $$\frac{q+1}{2} \leq \frac{q}{2 - (1/n^2)}.$$
But the inequality $$\frac{q+1}{2} \leq \frac{q}{2 - (1/n^2)}$$ is equivalent to $$2q + 2 - \frac{q + 1}{n^2} \leq 2q,$$ which in turn, is equivalent to $$n^2 \leq \frac{q+1}{2}.$$
This last inequality implies that $$n^2 \leq \frac{q+1}{2} < q < q^2,$$ from which it follows that $k = 1$.
We therefore have the true implication $$k \neq 1 \implies k = 1.$$
This means that $k = 1$ holds.
Here then is my:
QUESTION: Does this "proof" hold water? If it does not, where is the error, and can it be mended so as to produce a valid argument?
Edited in response to a comment from Bill Dubuque - 07/07/2022
So, basically my question is whether the inequality $$\frac{q+1}{2} \leq \frac{q}{2-(1/n^2)}$$ is indeed equivalent to $k \neq 1$, though I think this inquiry is fairly obvious from the context.
What you have written is a valid (assuming all of the statements you quote are correct) proof that $\frac{q+1}{2} \leq \frac{q}{2-1/n^2}$ is always false. The value of $k$ is irrelevant.
(And so it is not surprising that you can 'deduce' $k=1$ from the assumption that $\frac{q+1}{2} \leq \frac{q}{2-1/n^2}$ - false statements imply everything!)