Prove any integer multiple of a divisor rich number is divisor rich itself?

Question

A positive integer $n$ is divisor rich if the sum of its proper divisors is strictly greater than $n$. For instance, $10$ is not divisor rich because its proper divisors are $1, 2,$ and $5,$ which add to 8. Also, squares don't have the same proper divisor twice, only once. Also, the number $n$ itself doesn't count as a proper divisor, just like in the example with $10$. Is there any algebraic proof to this?

Thanks!

Answer 1

If $n$ has proper divisors $d_1, \ldots, d_k$ with $d_1 + \ldots + d_k > n$, then the proper divisors of $mn$ include $md_1, \ldots, m d_k$ with ...

Answer 2

Hint: If $d$ is a proper divisor of $n$, then $dk$ is a proper divisor of $nk$.

Answer 3

If $n=p_1^{a_1}p_2^{a_2}\dots p_r^{a_r}$ then $\frac{\sigma(n)}{n}=\prod\limits_{j=1}^r \frac{p^{a_j+1}-1}{p^j(p-1)}$.

Clearly each factor is an increasing function with respect to $a_j$ and every factor is greater than or equal to $1$. So increasing the number of factors or increasing the exponents only makes the fraction larger.

Answer 4

If a divisor rich n is multiplied by d >= 2, then we can assume d is prime, otherwise multiply by each prime factor of d in turn, and each product is divisor rich.

The proper factors of dn are n, each proper factor of n multiplied by d, and the proper factors of n which are not divisible by d. So the sum of divisors is at least (d+1)n. If the sum of proper divisors before was cn for some c >= 1, then it is now more than (c*d + 1)*n = (c + 1/d) * dn.