About the sigma function and an interesting inequality.

Question

Is it true $\sigma(A)$/$\sigma(B)$ > = (A/B) ; given B divides A ?

Answer 1

I assume that the question refers to the standard sum-of-divisors function $\sigma_1(n)$. If $B$ divides $A$, we can write $A=Bk$ with $k$ integer. We clearly have $\frac{\sigma_1(A)}{\sigma_1(B)}=\frac{A}{B}$ when $k=1$ and $A=B$.

On the other hand, when $k$ contains a single prime factor (i.e., $k$ is a prime number), the set of divisors of $A$ includes all divisors of $B$, plus all these multiplied by $k$. Thus, we have

$$\sigma_1(A)=(k+1)\sigma_1(B)$$

and then

$$\frac{\sigma_1(A)}{\sigma_1(B)}=k+1>\frac{A}{B}$$

It is not difficult to show that this inequality is also true when $k$ contains more than one factor, as in this case the set of divisors of $A$ includes all divisors of $B$, plus all these multiplied by each possible combination of factors of $k$.