Let $a_1, a_2, \dots, a_{\delta(n)}$ be the positive divisors of $n$ in increasing order. We obtain
$$\sigma(n) = a_1 + a_2 + \dots + a_{\delta(n)} \ge 1 + 2 + \dots + \delta(n) = \frac{\delta(n) \cdot (\delta(n)+1)}{2} > \frac{\delta(n)^2}{2}.$$

For every positive divisor $d$ of a positive integer $n$, $d + \frac{n}{d} \ge 2\sqrt{d \cdot \frac{n}{d}} = 2\sqrt{n}$. Thereby if $d \ne \sqrt{n}$, the inequality is strict. If $n$ is not a perfect square, then by adding all these inequalities for divisors $d < \sqrt{n}$ leads to
$$\sigma(n) > \frac{\delta(n)}{2} \cdot 2\sqrt{n} = \delta(n)\sqrt{n},$$
since every divisor of $n$ occurs exactly once in a pair $(d, \frac{n}{d})$. If $n$ is a perfect square, then analogously
$$\sigma(n) \ge \frac{\delta(n)-1}{2} \cdot 2\sqrt{n} + \sqrt{n} = \delta(n)\sqrt{n}.$$
On the other hand, if $n$ is not a perfect square then the number of pairs $(d, \frac{n}{d})$ where $d < \sqrt{n}$, is less than $\sqrt{n}$, whence $\sqrt{n} > \frac{\delta(n)}{2}$. If $n$ is a perfect square then the number of such pairs is at most $\sqrt{n}-1$, which gives $\sqrt{n}-1 \ge \frac{\delta(n)-1}{2}$ implying $\sqrt{n} > \frac{\delta(n)}{2}$ again. Hence,
$$\sigma(n) \ge \delta(n)\sqrt{n} > \delta(n) \cdot \frac{\delta(n)}{2} = \frac{\delta(n)^2}{2}.$$