Inverse sum of amicable numbers

Let $m , \; n$ be a pair of amicable numbers. Prove that:

$$\mathbf{\left ( \sum_{d \mid m}\frac{1}{d} \right )^{-1}+ \left ( \sum_{d \mid n} \frac{1}{d} \right )^{-1}=1}$$

Solution

We are beginning with a very well known exercise that we are taking it as a lemma:

Lemma: It holds that:

$$\mathbf{\sum_{d \mid m} \frac{1}{d}= \frac{\sigma(m)}{m}}$$

Proof:
Since the function $\frac{1}{n}$ is multiplicative , it follows that the function
$$F(n)=\sum_{d \mid n} \frac{1}{d}$$

is also multiplicative. Thus, the function $G(n)=\frac{\sigma(n)}{n}$ is also multiplicative. If $p$ is prime and $a$ is a natural number greater or equal to $1$, we have that:

$$F(p^a)=1+\frac{1}{p}+\cdots + \frac{1}{p^a}= \frac{1+p+\cdots+p^a}{p^a}= \frac{\sigma(p^a)}{p^a}=G\left ( p^a \right )$$

and so these two functions are equal.

 
 Back to our problem we have successively:

\begin{align*}
\mathbf{\left ( \sum_{d \mid n} \frac{1}{d} \right )^{-1} + \left ( \sum_{d \mid m} \frac{1}{d}\right )^{-1}} &=\mathbf{\frac{n}{\sigma(n)}+ \frac{m}{\sigma(m)}} \\
 &\!\overset{(*)}{=}\mathbf{\frac{n}{n+m} + \frac{m}{m+n}} \\
 &= \mathbf{\frac{n+m}{n+m}} \\
 &=\mathbf{1}
\end{align*}

$(*)$ Since the two numbers are amicable it holds that:

$$\mathbf{\sigma(m)=m+n=\sigma(n)}$$