Let $D(u)$ be the density of positive integers $n$ with $\sigma(n) \leq un$. (Let's take for granted that this density indeed exists; if you want, feel free to use that it is continuous, as well.) Assuming without proof (if you wish to use it) $\sum_{n\leq x} \frac{\sigma(n)}{n} \sim \frac{\pi^2}{6}x$, show that $D(u) \to 1$ as $u\to \infty$.
I do not see how to nicely relate $\sum_{n\leq x, \sigma(n) \leq un} 1$ with $\sum_{n\leq x} \frac{\sigma(n)}{n}$ - Partial Summation does not really help here, or at least I do not see how. Any help appreciated!
Let $D_x(u) = \sum_{n\leq x,\sigma(n)\leq un} 1$, so that $D(u)=\lim_{x\to\infty}D_x(u)/x$. Now split up the sum: \begin{align*} \sum_{n\leq x}\frac{\sigma(n)}{n}&=\left(\sum_{n\leq x,\sigma(n)\leq un}+\sum_{n\leq x,\sigma(n)>un}\right)\frac{\sigma(n)}{n}\\ &\geq\sum_{n\leq x,\sigma(n)\leq un}\frac{\sigma(n)}{n}+u\sum_{n\leq x,\sigma(n)>un}1\\ &=\sum_{n\leq x,\sigma(n)\leq un}\frac{\sigma(n)}{n}+u(\lfloor x\rfloor-D_x(u))\\ \end{align*} It follows that: \begin{align*} \frac{1}{x}\sum_{n\leq x}\frac{\sigma(n)}{n}\geq \frac{1}{x}\sum_{n\leq x,\sigma(n)\leq un}\frac{\sigma(n)}{n}+u(1-\frac{D_x(u)}{x}+O(1/x)) \end{align*} Now let $x\to\infty$ and then $u\to\infty$.