Jtdylktuy

Let $sigma(x)$ denote the sum of the divisors of $x$. Denote the deficiency of $x$ by $D(x)=2x-sigma(x)$.

If $N$ is odd and $sigma(N)=2N$, then $N$ is called an odd perfect number. Euler showed that an odd perfect number, if one exists, must necessarily have the so-called Eulerian form $N=q^k n^2$, where $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.

Since $k equiv 1 pmod 4$ and $q$ is prime, we obtain
$$sigma(q^k) = frac{q^{k+1}-1}{q-1} = bigg(frac{q^{frac{k+1}{2}}-1}{q-1}bigg)bigg(q^{frac{k+1}{2}}+1bigg).$$

But we can rewrite the first factor as
$$frac{q^{frac{k+1}{2}}-1}{q-1} = frac{q^{frac{k-1}{2}+1}-1}{q-1} = sigma(q^{frac{k-1}{2}}),$$
so that we have
$$sigma(q^k) = sigma(q^{frac{k-1}{2}})bigg(q^{frac{k+1}{2}}+1bigg).$$

Note that the divisor-sum function $sigma$ satisfies $sigma(yz) leq sigma(y)sigma(z)$, where equality holds if and only if $gcd(y,z)=1$. Setting $y=z=q^{frac{k-1}{2}}$ and noting that $gcd(y,z)=q^{frac{k-1}{2}}>1$ (unless $k=1$), we obtain
$$sqrt{sigma(q^{k-1})} < sigma(q^{frac{k-1}{2}}).$$

Now, it is known that
$$frac{2n^2}{sigma(q^k)}=frac{sigma(n^2)}{q^k}$$
from which it follows that
$$frac{D(n^2)}{sigma(q^{k-1})}=frac{2n^2 - sigma(n^2)}{sigma(q^k) - q^k}=frac{2n^2}{sigma(q^k)},$$
whereby we obtain
$$frac{2n^2}{D(n^2)}=frac{sigma(q^k)}{sigma(q^{k-1})}=frac{sigma(q^{frac{k-1}{2}})bigg(q^{frac{k+1}{2}}+1bigg)}{sigma(q^{k-1})}>frac{sqrt{sigma(q^{k-1})}bigg(q^{frac{k+1}{2}}+1bigg)}{sigma(q^{k-1})}=frac{q^{frac{k+1}{2}}+1}{sqrt{sigma(q^{k-1})}}.$$
This implies that
$$frac{n^2}{D(n^2)}>frac{q^{frac{k+1}{2}}+1}{2sqrt{sigma(q^{k-1})}}.$$

We now claim the following proposition.

If $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then
$$frac{sigma(n^2)}{n^2} leq frac{2q}{q+1}.$$
Equality holds if and only if $k=1$.

Proof

$$sigma(q^k)sigma(n^2)=sigma(q^k n^2)=2q^k n^2$$
$$frac{q+1}{q}=frac{sigma(q)}{q} leq frac{sigma(q^k)}{q^k}=frac{2n^2}{sigma(n^2)}$$
implying
$$frac{sigma(n^2)}{n^2} leq frac{2q}{q+1}.$$
Equality holds if and only if $k=1$.

From the inequality
$$frac{sigma(n^2)}{n^2} leq frac{2q}{q+1}$$
we obtain
$$(q+1)sigma(n^2) leq 2qn^2$$
which implies that
$$2n^2 leq 2qn^2 + 2n^2 - qsigma(n^2) - sigma(n^2) = {2n^2}(q+1) - (q+1)sigma(n^2) = (q+1)(2n^2 - sigma(n^2)) = (q+1)D(n^2)$$
from which it follows that
$$frac{n^2}{D(n^2)} leq frac{q+1}{2}.$$
Equality holds if and only if $k=1$.

Now assume that $k>1$. It follows from our previous considerations in this post that
$$frac{q+1}{2}>frac{n^2}{D(n^2)}>frac{q^{frac{k+1}{2}}+1}{2sqrt{sigma(q^{k-1})}}$$
which implies that
$$sqrt{sigma(q^{k-1})}>frac{q^{frac{k+1}{2}}+1}{q+1}$$
from which it follows that
$$frac{q^k - 1}{q - 1}=sigma(q^{k-1})>bigg(frac{q^{frac{k+1}{2}}+1}{q+1}bigg)^2.$$
Since $k>1$ and $k equiv 1 pmod 4$, then the lowest possible value of $k$ is $5$. Plugging this as a test value, we obtain
$$frac{q^5 - 1}{q - 1}=q^4 + q^3 + q^2 + q + 1$$
and
$$bigg(frac{q^3 + 1}{q + 1}bigg)^2 = (q^2 - q + 1)^2 = q^4 - 2q^3 + 3q^2 - 2q + 1$$
so that we get
$$q^4 + q^3 + q^2 + q + 1 > q^4 - 2q^3 + 3q^2 - 2q + 1$$
which simplifies to
$$3q^3 - 2q^2 + 3q = q(q(3q - 2) + 3) > 0$$
which is trivially true as it is known that $q geq 5$ (since $q$ is a prime satisfying $q equiv 1 pmod 4$). Indeed, by a lengthy (but doable) computation, it is possible to show that the inequality
$$frac{q^k - 1}{q - 1}=sigma(q^{k-1})>bigg(frac{q^{frac{k+1}{2}}+1}{q+1}bigg)^2$$
is trivial.

Here is my question:

The crux of the inequalities in this post is
$$sqrt{sigma(q^{k-1})} < sigma(q^{frac{k-1}{2}}).$$
Will it be possible to improve on this inequality, by a similar reasoning for the derivation of the inequality
$$bigg(frac{sigma(n^2)}{n^2}bigg)^{frac{ln(4/3)}{ln(13/9)}} < frac{sigma(n)}{n},$$
as detailed in this MSE question?

The conjecture in the comments is false. The following counterexample was communicated to me by Pascal Ochem. (He adds that the inequality
$$sqrt{sigma(q^{k-1})} < sigma(q^{frac{k-1}{2}})$$
cannot be improved much.)

With $q = 40009$ and $k = 5$:
$$(q^4+q^3+q^2+q+1)^{1/2} = 1600740085.875$$
$$q^2+q+1 = 1600760091$$
$$(q^4+q^3+q^2+q+1)^{log(6/5)/log(31/25)} = 4004779171150154.163$$