Preamble: This question is an offshoot of this earlier MSE post.
Consider a hypothetical odd perfect number $N=p^k m^2$ with special/Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$, and let $\varphi(x)$ denote the Euler's totient function of $x$.
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 $$\dfrac{240}{217\zeta(3)} < \dfrac{\sigma(N)\varphi(N)}{N^2} < 1,$$ where we have the rational approximation $$\dfrac{240}{217\zeta(3)} \approx 0.92008188672520641697672952496390495972393334311564395865.$$
Note that this result significantly improves on the classical (lower) bound(s) $$\dfrac{6}{{\pi}^2} < \dfrac{\sigma(n)\varphi(n)}{n^2} < 1,$$ which holds for all positive integers $n$, where we have the rational approximation $$\dfrac{6}{{\pi}^2} \approx 0.607927101854026628663276779258365833426152648.$$
As derived in the linked question, we have the bounds $$\dfrac{120}{217\zeta(3)} < \dfrac{\varphi(m)}{m} < \dfrac{5}{8},$$ where we have the rational approximation $$\dfrac{120}{217\zeta(3)} \approx 0.4600409433626.$$
Here is my initial inquiry:
Would it be possible to improve the lower bound for $\varphi(m)/m$ to $$\dfrac{1}{2} < \dfrac{\varphi(m)}{m}?$$
Note that it is not possible to have $$\dfrac{1}{2} = \dfrac{\varphi(m)}{m}$$ since this implies $$m=2\varphi(m),$$ which results in the contradiction "LHS is odd, RHS is even".
MY "ATTEMPT"
The classical (unconditional) upper bound for $$\dfrac{\sigma(n)\varphi(n)}{n^2} < 1$$ gives $$\dfrac{\sigma(n)}{n} < \dfrac{n}{\varphi(n)},$$ which holds for any positive integer $n$. Letting $n=m$ (the factor of our odd perfect number), we obtain $$\dfrac{\sigma(m)}{m} < \dfrac{m}{\varphi(m)}.$$ Now, the RHS of the last inequality is bounded from above by $2$ if $1/2 < \varphi(m)/m$. Hence, we get $$\dfrac{\sigma(m)}{m} < \dfrac{m}{\varphi(m)} < 2,$$ which does hold (since $m$ is deficient, being a proper factor of the (odd) perfect number $p^k m^2$).
CAVEAT
I am, however, aware that it may be possible to have $$\dfrac{\sigma(m)}{m} < 2 < \dfrac{m}{\varphi(m)},$$ whence the conclusion that I require does not hold.
Here is my final inquiry:
It might then be more fruitful to consider the cases $$\dfrac{\varphi(m)}{m} < \dfrac{1}{2}$$ and $$\dfrac{1}{2} < \dfrac{\varphi(m)}{m}$$ separately. Do you have any suggestions on how this can be done?