The square suggests Cauchy-Schwarz...

$\displaystyle \sum_{j=1}^n \frac{1}{j}=\sum_{j=1}^n \frac{\sqrt{\sigma(j)}}{j}\frac{1}{\sqrt{\sigma(j) }}$ $\displaystyle \leq \left(\sum_{j=1}^n \frac{\sigma(j)}{j^2}\right)^{1/2}\left(\sum_{j=1}^n\frac{1}{\sigma(j)}\right)^{1/2}$.

Then note that $\displaystyle \sum_{j=1}^n \frac{1}{\sigma(j)}\leq\sum_{j=1}^n\frac{1}{j}$ (if we order the values of $\displaystyle \sigma(1),\ldots,\sigma(n)$ as $\displaystyle a_1<\ldots< a_n$, then by induction we have $\displaystyle a_j\geq j$ since the sequence is strictly increasing).

Plugging this inequality in the above one and simplifying yields the inequality we are looking for.

Assume there is equality. Then there is equality in Cauchy-Schwarz inequality, which implies that, for some $\displaystyle \lambda\geq 0$, for $\displaystyle j=1,\ldots,n$, $\displaystyle \frac{\sqrt{\sigma(j)}}{j}=\frac{\lambda}{\sqrt{\s igma(j)}}$ hence $\displaystyle \sigma(j)=\lambda j$. Since $\displaystyle \sigma(1)$ is integer-valued, we must have $\displaystyle \lambda\in\mathbb{N}^*$. And there is also equality in the other inequality : $\displaystyle \sum_{j=1}^n\frac{1}{j}=\sum_{j=1}^n\frac{1}{\sigm a(j)}$, which entails $\displaystyle \lambda=1$ immediately. Thus, $\displaystyle \sigma(j)=j$ for $\displaystyle j=1,\ldots,n$.