配对的整数因子

Oftentimes, it is useful to think of factors in pairs. The best way to illustrate this is swith an example.

Let $\tau(n)$ be the number of divisor of $n$. Prove that $\prod\limits_{d\mid n,\,d>0}d=n^{\tau(n)/2}$

Note that if $d$ is in the product, then we know $d\mid n$, and there exists an integer $k$ such that $dk=n$. Thus, $k\mid n$, and $k=\frac{n}{d}$ is also in the sum. Note that this creates a bijection from the set of divisors to itself. So, let's square the equation so we have two of the products. $$\left[\prod\limits_{d\mid n,\,d>0}d\right]\left[\prod\limits_{d\mid n,\,d>0}d\right]=n^{\tau(n)}.$$ We know that one of the products can be rewritten. $$\left[\prod\limits_{d\mid n,\,d>0}d\right]\left[\prod\limits_{d\mid n,\,d>0}\frac{n}{d}\right]=n^{\tau(n)}.$$ Combining the products, $$\left[\prod\limits_{d\mid n,\,d>0}n\right]=n^{\tau(n)}.$$ We know this is true, because there are $\tau(n)$ divisors, and we are multiplying it that many times. So, since this is true and all steps are reversible, the first equation is true.

Let $\sigma(n)$ be sum of divisors of $n$, $\tau(n)$ be number of divisor of $n$. Prove that for:

(a) Any $n \ge 7$ then $\sigma(n)<n \ln n$.

(b) Any $n \ge 1$ then $\dfrac{\sigma(n)}{\tau(n)} \ge \sqrt{n}$.