算术和乘法函数的进一步介绍

The M$\ddot{\textbf{o}}$bius function $\mu(n)$ is defined as follows:

$\bullet$ $\mu(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.

$\bullet$ $\mu(n) = -1$ if $n$ is a square-free positive integer with an odd number of prime factors.

$\bullet$ $\mu(n) = 0$ if $n$ has a squared prime factor.

---

This function has many interesting properties, which will be discusses in the official post when algebra and number theory get launched.

Prove that, for a multiplicative function $f$, $$\sum_{d|n}\mu(d)f(d)=\prod_{p|n}(1-f(p)).$$

The Mobius function is denoted as $\mu(n)$ and counts the number of prime factors in n, when n has a square factor (obviously not 1), $\mu(n)$ is 0, and when n has m many primes, $\mu(n)$ is equal to -1 to the power of m. It is useful in many abstract branches of number theory.

Prove that, $$\left|\sum_{d|n}\dfrac{\mu(d)\sigma(d)}{d}\right|\geq\dfrac{1}{n}$$