The question simply states
"Prove that any multiple of a perfect number is abundant."
Perfect number and Abundant number are each defined by $\sigma(n)=2n$ and $\sigma(n)>2n$
How to solve this question even the statement is quite simple?
My try:
First I let the perfect number $n$ be $n=2^{k-1}(2^k-1)$ then consider a multiple $m$ with various forms like $q, 2^\lambda q, (2^k-1)^aq, 2^\lambda(2^k-1)^aq$ but it takes me too long and yields nothing.
Then I suddenly thought that I need to consider odd perfect number (even if it doesn't exist at now time), so I throw out the form $n=2^{k-1}(2^k-1)$ and simply consider $\sigma(n)=2n$ , then I let $m$ be the multiple and successfully proved that when $\gcd(m,n)=1$ then the inequality $\sigma(mn)>2mn$ indeed hold, but I'm losing my way when $\gcd(m,n)=d>1$, or let me be more precise, I don't know how to continue with $\sigma(mn)$ when they are not relatively prime.
Thanks to @gandalf61 ! $$\sigma (mn)\geq\sigma(m)\sigma(n)=\sigma(m)(2n)>2mn$$ So this completes the proof.