In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213

A results is:

If the Riemann hypothesis is True and $n ≥ 5041$
$\frac{\sigma(n)}{n} < e^\gamma \ln \ln (n)$

We also know that $e^\gamma < e$ , Now my question here is  :

Question: Without using the Riemann hypothesis, is it possible to show that: $\frac{\sigma(n)}{n} < e \ln \ln (n)$ ?

In Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213 (pdf) we find the following result:

If the Riemann hypothesis is true and $n ≥ 5041$, $\frac{\sigma(n)}{n} < e^\gamma \ln \ln (n)$

We also know that $e^\gamma < e$. Now my question here is:

Question: Without using the Riemann hypothesis, is it possible to show that $\frac{\sigma(n)}{n} < e \ln \ln (n)$ ?

In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213

In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213

A results is:

If the Riemann hypothesis is True and $n ≥ 5041$
$\frac{\sigma(n)}{n} < e^\gamma \ln \ln (n)$

We also know that $e^\gamma < e$ , Now my question here is :

Question: Without using the Riemann hypothesis, is it possible to show that: $\frac{\sigma(n)}{n} < e \ln \ln (n)$ ?

In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213

A results is:

If the Riemann hypothesis is True and $n ≥ 5041$
$\frac{\sigma(n)}{n} < e^\gamma \ln \ln (n)$

We also know that $e^\gamma < e$ , Now my question here is :

Question: Without using the Riemann hypothesis, is it possible to show that: $\frac{\sigma(n)}{n} < e \ln \ln (n)$ ?

In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213