About the 'sigma' function.

Question

Is it true that if $n$ divides $m$ , $\sigma(\frac mn) \leq \frac{\sigma(m)}n$. If so this has a bearing on counterexamples to Robin's inequality.

Answer 1

Write $m=kn$ then we show that $$n\sigma(k)\leq\sigma(m)$$ for every positive n we have $n\leq\sigma(n)$ then $$n\sigma(k)\leq\sigma(n)\sigma(k)\ \ \ \ (*)$$ if $gcd(n,k)=1$ $\sigma(n)\sigma(k)=\sigma(m)$ because $\sigma$ is a multiplicative function.

If $gcd(n,k)=d\neq1$ we can write $n=ds \ and\ k=dt$ now for $(*)$ we have that $$n\sigma(k)\leq\sigma(n)\sigma(k)\leq\sigma(d)\sigma(s)\sigma(d)\sigma(t)$$ we want now that $$\sigma(d)\sigma(d)\sigma(s)\sigma(t)\leq\sigma(d^2)\sigma(s)\sigma(t)=\sigma(m)$$ i.e. $$\sigma(d)\sigma(d)\leq\sigma(d^2)$$ for multiplicativity of $\sigma$ we can prove this only for $d=p^a$ where p is prime and $a\geq1$ $$\sigma(p^a)=\frac{p^{a+1}-1}{p-1}$$ so we have to prove $$\left(\frac{p^{a+1}-1}{p-1}\right)^2\leq\frac{p^{2a+1}-1}{p-1}$$ and this it's true.

Answer 2

Let $k=\frac mn$, and let $1=d_1\lt d_2\lt\dots\lt d_r=k$ be all the positive divisors of $k$. Then $nd_1\lt nd_2\lt\dots\lt nd_r$ are distinct positive divisors of $nk$, and so $\sigma(m)=\sigma(nk)\ge nd_1+nd_2+\cdots+nd_r=n(d_1+d_2+\cdots+d_r)=n\sigma(k)=n\sigma(\frac mn)$.