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.

Why is it that we cannot go below $1$ for the constant $K$ by checking the inequality for more cases by hand?

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

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 $1 \leq n \leq 106$.

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.

Why is it that we cannot go below $1$ for the constant $K$ by checking the inequality for more cases by hand?

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.

Why is it that we cannot go below $1$ for the constant $K$ by checking the inequality for more cases by hand?