Robin's criterion, Goldbach's conjecture and upper bound for $r_{0}(n)$

This question is a follow-up to both About Goldbach's conjecture and Question in Proof of Hardy Ramanujan theorem about $\omega(n) =\sum_{p|n} 1$.

Can one derive from Robin's criterion for RH an inequality of the form $$\sigma(n)\ll n\omega(n)$$? Moreover, as Carl Pomerance told me in a private communication that the quantity $$r_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$$ is expected to fulfill $$r_{0}(n)\ll\log^{2}n\log\log n$$, can one get from this criterion an upper bound for $$r_{0}(n)$$ of the form $$r_{0}(n)\ll\frac{\sigma(n)}{N_{2}(n)}$$ where $$N_{2}(n)=\sharp\{0 under RH? Finally, whenever $$n$$ is a perfect number, does $$\sigma(n)=2n$$ imply (always assuming RH) that $$r_{0}(n) for some positive constant $$C$$?

The asymptotic $$\sigma(n)\ll n\omega(n)$$ is true and has nothing to do with RH or Robin's criterion.

Let us rewrite it as $$\frac{\sigma(n)}{n}\ll\omega(n)$$. By writing $$n=\prod_{i=1}^k p_i^{e_i}$$, where $$p_1<\dots are primes and $$e_i>0$$, we want to show $$\prod_{i=1}^k\frac{\sigma(p_{e_i}^{e_i})}{p_{e_i}^{e_i}}\ll k.$$ But $$\frac{\sigma(p^k)}{p^k}=1+\frac{1}{p_i}+\dots+\frac{1}{p_i^{e_i}}<1+\frac{1}{p_i-1}\leq 1+\frac{1}{i}$$, so $$\frac{\sigma(n)}{n}=\prod_{i=1}^k\frac{\sigma(p_{e_i}^{e_i})}{p_{e_i}^{e_i}}\leq\prod_{i=1}^k\left(1+\frac{1}{i}\right)=k+1.$$

I have no idea about the quantities related to Goldbach conjecture but I strongly doubt RH can prove anything like what you ask for.