I hope 2015 doesn't start with gross nonsense question for me.
Robins equivalence for RH is
$$ \frac{\sigma(n)}{n \log\log n} < e^{\gamma}\qquad(1)$$
for $n \ge 5041$. We have $ \limsup \frac{\sigma(n)}{n \log\log n} = e^{\gamma}$.
Lagarias equivalence is
$$ \sigma(n) < H_n + \exp(H_n) \log(H_n) \qquad(2)$$
for $n \ge 2$ and $H_n$ the harmonic numbers.
(1) and (2) are closely related.
According to "Sharp Bounds for the Harmonic Numbers"
$$ \frac{1}{2n +\frac{1}{1-\gamma} - 2} \le H_n - \log n - \gamma$$
and a simpler bound for $n$ large enough is $\log{n} + \gamma \le H_n $
Assume $\log\log{n} > 1$. Substituting the simpler bound in (2) and dividing by $n \log{(\log{n}+\gamma)}$:
$$\frac{\sigma(n)}{n\log{(\log{n}+\gamma})} < e^{\gamma}+\frac{\log{n}+\gamma}{n\log{(\log{n}+\gamma)}} \qquad (3)$$
Since we substitued the lower bound for $H_n$, (3) implies RH.
The LHS of (3) is strictly smaller than the LHS of (1) since $ \frac{\sigma(n)}{n \log\log n} > \frac{\sigma(n)}{n\log{(\log{n}+\gamma})}$ and the RHS of (3) is larger than RHS of (1) since $\frac{\log{n}+\gamma}{n\log{(\log{n}+\gamma)}}$ is positive.
Because of this (3) might be easier to prove and $(1) \implies (3)$.
It can't happen (3) to be true and (1) false, since in this case RH will be both true and false, so $(1) \iff (3)$.
Is it true that (3) is equivalent to RH?
In case of positive answer:
Define $G(n)=\frac{\sigma(n)}{n \log\log n}$.
(1) and (3) are logically equivalent though technically different. Does this imply $G(n)$ can't take non-trivial values/ranges because of contradiction?