Here, we prove the following proposition.

**Theorem.**If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$$

implies that

$$k=1$$

and

$$q=5.$$

**Proof.**Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

Since $N$ is perfect, we have

$$2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \left(\frac{q^{k+1} - 1}{q - 1}\right)\cdot\sigma(n^2).$$

It follows that

$$\frac{\sigma(n^2)}{n^2} = \frac{2{q^k}\left(q - 1\right)}{q^{k+1} - 1} = \frac{2q^{k+1} - 2q^k}{q^{k+1} - 1} = \frac{2q^{k+1} - 2}{q^{k+1} - 1} - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right)$$

$$= 2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right).$$

By assumption, we have

$$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}.$$

This inequality is equivalent to

$$2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right) \geq 2 - \frac{5}{3q},$$

which, in turn, is equivalent to

$$6q\left(q^k - 1\right) \leq 5(q^{k+1} - 1).$$

This last inequality simplifies to

$$q^{k+1} - 6q + 5 \leq 0,$$

which cannot be true when $k > 1$. Thus, we know that $k=1$.

Hence, we have

$$q^2 - 6q + 5 \leq 0.$$

But this inequality implies that

$$1 \leq q \leq 5,$$

which, together with $q \geq 5$, implies that $q=5$.

**QED.**