Proof regarding Robin's inequality (RI). Let $\sigma$ be the divisor sum function, $\gamma$ the Euler-Mascheroni constant and $n>5040$. Robin showed that if the inequality$$\displaystyle \sigma(n)<e^{\gamma}n\log\log n$$ ever fails, it does infinitely often. It is quite intuitive that infinitely many of the counterexamples (if not all) would be superabundant, i.e. natural numbers $a$ such that $\displaystyle \frac{\sigma(a)}{a}>\frac{\sigma(b)}{b}$ for all $b<a$. 
My proof, which I'd like to have verified:

Let $SA_k$ be the $k$-th superabundant number. Then assume, without loss of generality, 
$$
\left\{ 
\begin{array}{c}
\sigma(SA_k)<e^{\gamma}SA_k\log\log SA_k \\ 
\sigma(m)\ge e^{\gamma}m \log \log m \\ 
SA_l < SA_k<m<SA_{k+1}
 \ , \end{array}
\right. 
$$
where $SA_l$ is the largest superabundant counterexample.
So we must have $$\displaystyle \frac{\sigma(m)}{\sigma(SA_k)}>\frac{m \log\log m}{SA_k\log\log SA_k} \\ \ \ \ \ \ \ \frac{\sigma(m)}{m \log\log m} >\frac{\sigma(SA_k)}{SA_k\log\log SA_k}. \ \ \ \ \ \  (1)$$
But since $m$ is between two consecutive superabundant numbers, it is not superabundant itself, hence it is by definition $\displaystyle \frac{\sigma(m)}{m}\le\frac{\sigma(SA_k)}{SA_k}$. Given that $ \log\log m >\log\log SA_k$, we easily find $$\displaystyle \frac{\sigma(m)}{m \log\log m} <\frac{\sigma(SA_k)}{SA_k\log\log SA_k},$$ contradicting $(1)$. As a result, we have no counterexamples $> SA_l$, which is an absurdity. $ \ \ \ \ \ \ \ \ \ \  \square$

Any comment, suggestion for improvement or alternative proof would be highly appreciated.
 A: Looks right, although you have thrown in subscripts and extra letters that are not needed. Note that Akbary and Friggstad pointed out in 2009 that the first counterexample to Robin would be superabundant; given your argument, the reason that not all counterexamples need be superabundant is that a superabundant counterexamples can be followed by a trickle of not-quite-superabundant counterexamples that are smaller than the next superabundant number.
Shorter: $A$ is largest superabundant counterexample. There are infinitely many counterexamples (if any), so there is a counter example $n$ larger than some superabundant number $B$ that is not a counterexample, that is consecutive superabundant $B,C $ with $B < n < C.$ We do not really care about $C,$ can also forget $A.$ 
We have, since we are assuming $n$ is not superabundant, indeed is smaller than the very next SA number,
$$ \frac{\sigma(n)}{n} \leq \frac{\sigma(B)}{B}.   $$ As $B$ is not a counterexample,
$$  \frac{\sigma(B)}{B \log \log B} < e^\gamma.   $$ Since $n > B,$ we also have
$$ \frac{1}{\log \log n} <  \frac{1}{\log \log B}. $$ Combine with $ \frac{\sigma(n)}{n} \leq \frac{\sigma(B)}{B}   $ to arrive at 
$$   \frac{\sigma(n)}{n \log \log n} < \frac{\sigma(B)}{B \log \log B} < e^\gamma.   $$
This contradicts $n$ being a counterexample, so, if there are any counterexamples, infinitely many are superabundant.
Note that Robin showed in his section 3 that at least one colossally abundant number is a counterexample if there are any. I'm not sure it is known about infinitude of CA counterexamples, if any.
EEDDIITT: I once did something computational here with CA numbers, incredible amount of work for a disappointing outcome 
 How would I find a number where $\sum_{d\mid n}d > 100n$?  The conclusion is that RH is not going to be decided by something that can be done on a computer. 
