Rather than multiplying, we sum $\forall i \in {1, 2, \ldots \omega(N)}$ to get:
$$\sum_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le \frac{2\omega(N)}{3}$$
Following Nielsen, we know that (for lack of an "effective" upper bound for $\omega(N)$):
$$\inf\left({\frac{2\omega(N)}{3}}\right) \ge 6$$
Now multiplying, $\forall i \in {1, 2, \ldots \omega(N)}$ we get:
$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} = \frac{{\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}\displaystyle\prod_{j = 1}^{\omega(N)}}{{\sigma({q_j}^{\beta_j})}}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{N{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}{\displaystyle\prod_{j = 1}^{\omega(N) - 1}{N}} = \displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}}$$
But:
$$\displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}} = \displaystyle\frac{2N}{N^{\omega(N) - 1}} = \displaystyle\frac{2}{N^{\omega(N) - 2}} = 2{N^{2 - \omega(N)}}$$
Since $7 \le \omega(N) - 2 = r$, it follows that $N^7 < N^r$$N^7 \le N^r$, which gives:
$$N^{-r} < N^{-7}$$$$N^{-r} \le N^{-7}$$
Consequently:
$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} < 2N^{-7} = \displaystyle\frac{2}{N^7}$$$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le 2N^{-7} = \displaystyle\frac{2}{N^7}$$
Taking logarithms of both sides of the last inequality:
$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) < \log(2) - 7\log(N)$$$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) \le \log(2) - 7\log(N)$$
This is as far as I could go, using only elementary notions that are familiar to me. I will let you guys know if I discover anything else in the coming days.
Post-Edit:
I missed it!
$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) = \displaystyle\sum_{j = 1}^{\omega(N)}{\left(\log({q_j}^{\beta_j}) + \log(\sigma({q_j}^{\beta_j})) - \log(N)\right)}$$ $$= \displaystyle\sum_{j = 1}^{\omega(N)}{\log({q_j}^{\beta_j})} + \displaystyle\sum_{j = 1}^{\omega(N)}{\log(\sigma({q_j}^{\beta_j}))} - \displaystyle\sum_{j = 1}^{\omega(N)}{\log(N)}$$ $$= \log(\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}) + \log(\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j})}) - \log(\displaystyle\prod_{j = 1}^{\omega(N)}{N})$$ $$= \log(N) + \log(\sigma(N)) - \log(N^{\omega(N)}) \le \log(2) - 7\log(N)$$
which implies that:
$$8\log(N) + \log(\sigma(N)) - \omega(N)\log(N) \le \log(2)$$
Finally, we have:
$$\log(\sigma_{1}(N)) \le \log(2) + \left(\omega(N) - 8\right)\log(N)$$
Equality holds if and only if $\omega(N) = 9$.