Is it possible to improve on these bounds for $\frac{\varphi(n)}{n}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  Denote the Euler-totient function of the positive integer $x$ by $\varphi(x)$, and the classical sum of divisors of $x$ by $\sigma(x)=\sigma_1(x)$.
MOTIVATION
From the following source:
Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$
However, we also have
$$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$
Notice that
$$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$
Therefore, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$
and
$$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$
which implies that
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$
WolframAlpha gives the rational approximation
$$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$
Harking back from this answer to a closely related question, we have the following proposition:

THEOREM A: Let $q^k n^2$ be an odd perfect number with special prime $q$.  Then we have the following implication:
$$\frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5.$$


PROOF: Let $q^k n^2$ be an odd perfect number with special prime $q$, and suppose that
$$\frac{\varphi(n)}{n} > \frac{1}{2}.$$
From the equation and lower bound for $\varphi(N)/N$
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$
and the equation
$$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$
we get the lower bound
$$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$
This implies that we have the upper bound
$$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$
which can be solved using WolframAlpha, yielding the upper bound
$$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$
from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.  QED

Note that the estimate
$$\frac{\varphi(n)}{n} > \frac{1}{2}$$
was used to justify
$$0 < q < \frac{x}{x-1}.$$

We now endeavor/attempt to prove the estimate
$$\frac{\varphi(n)}{n} > \frac{1}{2}.$$
To do so, consider the equation
$$\frac{2q}{q+1}=\frac{n}{\varphi(n)}.$$
(This is not random guesswork - it emanates from considering the inequalities
$$\frac{\sigma(n^2)}{n^2} < \frac{n^2}{\varphi(n^2)} = \frac{n}{\varphi(n)}$$
$$\frac{\sigma(n^2)}{n^2} = \frac{2q^k}{\sigma(q^k)} \leq \frac{2q}{q+1},$$
and determining whether the right-hand sides are equal.)
Since $q \equiv 1 \pmod 4$, then $(q+1)/2$ is an integer, so that we may rewrite
$$\frac{2q}{q+1}=\frac{n}{\varphi(n)}$$
as
$$q\varphi(n)=\bigg(\frac{q+1}{2}\bigg)\cdot{n}.$$
Since $\gcd(q,n)=1$, then $n \mid \varphi(n)$.  This contradicts
$$\frac{1}{2}<\frac{\varphi(n)}{n}=\frac{q+1}{2q}=\frac{1}{2}+\frac{1}{2q} \leq \frac{3}{5}.$$

An alternative proof goes by observing that
$$\gcd(q,n)=\gcd\Bigg(q,\frac{q+1}{2}\Bigg)=\gcd\Bigg(q,\frac{q+1}{2}\cdot{n}\Bigg)=1$$
which contradicts
$$q \mid \Bigg(\frac{q+1}{2}\cdot{n}\Bigg).$$

Thus, either
$$\frac{1}{2} < \frac{q+1}{2q} < \frac{\varphi(n)}{n} < \frac{5}{8}$$
or
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{q+1}{2q} < \frac{3}{5}$$
holds.

Here are our:
QUESTIONS

(1) Do you see a quick way to improve on these bounds, using the ideas we have given in this post, and possibly even more?


(2) If the bounds can be improved, can you show how?


(3) If the bounds cannot be improved, can you explain why?


Note:  This is not really an answer, just some thoughts that recently occurred to me, that would be too long to include in the Comments section.
Let $q^k n^2$ be an odd perfect number with special prime $q$.
It is known that
$$\varphi(x) < \sigma(x)$$
for all $x > 1$.  Therefore, we are sure that
$$\frac{\varphi(n)}{n} < \frac{\sigma(n)}{n} < \frac{\sigma(n^2)}{n^2} \leq \frac{2q}{q+1}.$$

Now, in the original post, we have proved that
$$\frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5.$$
Hence, we have
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q} \implies \frac{\varphi(n)}{n} > \frac{1}{2} \implies q = 5 \implies \frac{\varphi(n)}{n} > \frac{q+1}{2q} = \frac{5+1}{10} = \frac{3}{5}.$$
Therefore, the biconditional
$$q = 5 \iff \frac{\varphi(n)}{n} > \frac{3}{5}$$
holds, assuming the inequality
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q}$$
is true.

But we are sure that the inequality
$$\frac{\varphi(n)}{n} < \frac{2q}{q+1}$$
holds.
Assuming that
$$\frac{\varphi(n)}{n} > \frac{q+1}{2q}$$
is true, we obtain
$$\frac{q+1}{2q} < \frac{2q}{q+1}$$
$$(q + 1)^2 < 4q^2$$
which implies that
$$3q^2 - 2q - 1 > 0$$
$$(3q + 1)(q - 1) > 0$$
from which we finally conclude that
$$q > 1,$$
which is trivial.

Again, we know that the inequality
$$\frac{\varphi(n)}{n} < \frac{2q}{q+1}$$
holds.
Assuming that
$$\frac{\varphi(n)}{n} < \frac{q+1}{2q}$$
then we obtain
$$\Bigg(\frac{\varphi(n)}{n}\Bigg)^2 < \Bigg(\frac{2q}{q+1}\Bigg)\cdot\Bigg(\frac{q+1}{2q}\Bigg) = 1$$
from which we get
$$\frac{\varphi(n)}{n} < 1$$
which, again, is trivial.
