Question on coefficient of $\exp(H_n).\log(H_n)$ in Lagarias equivalence of RH In page 197, Equivalents of the Riemann Hypothesis Vol 1, the following statement caught my eye

There is an editorial comment in [102] that includes an observation by
the GCHQ Problem Solving Group. They contest that one could replace the
number 2 in inequality (7.94) by any constant $K > 1$ at the cost of having to
check more cases by hand when $K$ is close to one. An example is cited: when
$K = 1.2$ one needs to verify inequality (7.94) for $1 \leq n \leq 10^6$.

referring the inequality 7.94

$\sigma(n) \leq H_n +2\exp(H_n) \log(H_n)$, $n \geq 1$,

and the reference 102 being

[102] J. C. Lagarias and W. Janous, A generous bound for divisor sums: problem 10949, Amer. Math. Monthly 111 (2004), 264–265.

According to An Elementary Problem Equivalent to the Riemann Hypothesis by J. C. Lagarias, $K = 1$ is equivalent to RH. I understand the claim is not that RH can be proved by checking more cases by hand, but I want to understand what are the methods by which $K$ can be improved to any $1 + \epsilon$ by checking more cases? What are the barriers that prevent it from  proving that $K \leq 1$ by checking for more cases?
Edit: Edited my question with more details after looking at the answer by @Charles
 A: 1. By an inequality due to Robin (see (2.2) in Lagarias's paper),
$$\sigma(n)\leq e^\gamma n\log\log n+\frac{n}{\log\log n},\qquad n\geq 3.$$
By Lemma 3.1 in Lagarias's paper, we also know that
$$\exp(H_n)\log(H_n)\geq e^\gamma n\log\log n,\qquad n\geq 3.$$
Combining these two estimates, we infer that
$$\sigma(n)\leq\left(1+\frac{1}{(\log\log n)^2}\right)\exp(H_n)\log(H_n),\qquad n\geq 3.$$
The fraction on the right hand side tends to zero (effectively), confirming the observation by the GCHQ Problem Solving Group. Explicitly, given any $\varepsilon>0$, we have
$$\sigma(n)\leq(1+\varepsilon)\exp(H_n)\log(H_n),\qquad n\geq\exp\exp(\varepsilon^{-1/2}).$$
2. The barrier to prove Lagarias's inequality is the same as the barrier to prove the Riemann Hypothesis, since these two statements are equivalent. It is a difficult problem, and perhaps it is even undecidable from the ZFC axioms (of course most of us believe it is decidable).
A: The case $K=1$ is equivalent to the Riemann Hypothesis. [1] The GCHQ Problem Solving Group is not claiming that RH can be proved by checking a finite number of cases of this problem. I don't know what the barriers are that prevent it, though; I don't even know the method they used to extend $K$.
[1] Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis, American Mathematical Monthly 109 (2002), pp. 534–543.
