This question is more of a conjecture, but I am not posting it to mathoverflow, since I am not a professional mathematician and I do not know if it is research related: Let $n=p^{a} s$, $p$ be the smallest prime dividing $n$, $v_p(n) = a$, $n\equiv 1(2)$,$\sigma(n) = 2n$. Then we must have $p>2$. Let $m=2^a s $. Then we have $$\sigma(m)= \sigma(2^a)\sigma(s)= \frac{\sigma(2^a)\sigma(n)}{\sigma(p^a)}$$. Since $2<p$ we have $H(2^a) < H(p^a)$ where $H$ is the entropy of $n$ defined here: Entropy of a natural number. It follows that: $$H(n) = H(p^a s) = H(p^a)+H(s) > H(2^a) + H(s) = H(m) = \log(\sigma(m))-\frac{1}{\sigma(m)}\sum_{d|m}{d \log(d)} = \log(\frac{\sigma(2^a)\sigma(n)}{\sigma(p^a)})-\frac{1}{\sigma(m)}E(m)$$, where $E(k) = \sum_{d|k} {d \log(d)}$. Since $\log(x) \ge 1 - 1/x$ we have
$$ H(n) > H(m) = \log(\sigma(n))+\log(\frac{\sigma(2^a)}{\sigma(p^a)})-\frac{E(m)}{\sigma(m)}\ge\log(\sigma(n))+1-\frac{\sigma(p^a)}{\sigma(2^a)}-\frac{E(m)}{\sigma(m)}$$ which is equal to: $$\log(\sigma(n))+1-\frac{\sigma(n)}{\sigma(m)}-\frac{E(m)}{\sigma(m)}$$ If we can prove that, the last quantity is greater than or equal to $\log(\sigma(n))-\frac{E(n)}{\sigma(n)}= H(n)$, we would get the contradiction: $$H(n)>H(m) \ge H(n)$$ So we have to prove that: $$1 - \frac{\sigma(n)}{\sigma(m)}-\frac{E(m)}{\sigma(m)} \ge - \frac{E(n)}{\sigma(n)}$$ We have the following identities: $$E(m) = \sigma(2^a)E(s)+\sigma(s)E(2^a)$$ $$E(n) = \sigma(p^a)E(s)+\sigma(s)E(p^a)$$ Furthermore one can show that: $$E(p^a) = p \log(p) \frac{p^{a+1}-1-(a+1)p^a(p-1)}{(p-1)^2}$$ which can be applied to compute $E(2^a)$.
So my question is if somebody can help me prove the inequality above or show an "example" where it fails, if this is not too much to ask.
You require that $$\frac{E(p^a)}{\sigma(p^a)} \geq \frac{E(2^a)}{\sigma(2^a)}.$$
Assuming my calculations are correct, this inequality simplifies to: $$\frac{p\log p}{p - 1} - p\log p\frac{(a+1){p^a}}{p^{a+1} - 1} \geq 2\log 2 - 2\log 2\frac{(a+1){2^a}}{2^{a+1} - 1}$$
I asked WolframAlpha for the solutions of this inequality for $p$, it gave the following inequality plot:
This inequality plot can be interpreted to mean that the solutions are given by the sets of inequalities:
$$p \geq 2, a \leq 0$$
or
$$p \leq 2, a \geq 0.$$
Since neither of these two sets of inequalities satisfy the conditions of the problem, I am led to believe that your conjectured inequality does not hold.