An inequality for the sum-of-aliquot-divisors function

Question

In what follows, we shall assume that $a$ and $b$ are relatively prime. (That is, $\gcd(a,b)=1$ holds.)

It is known that the inequality $$\sigma(ab) \leq \sigma(a)\sigma(b)$$ holds for the sum-of-divisors function $\sigma(x)$.

It is also straightforward to show that the inequality $$D(ab) \leq D(a)D(b)$$ holds for the deficiency function $D(x)=2x-\sigma(x)$.

Here is an attempt towards proving that the corresponding inequality $$s(a)s(b) \leq s(ab)$$ holds for the sum-of-aliquot-divisors function $s(x)=\sigma(x)-x$:

$$s(ab)=\sigma(ab)-ab=\sigma(a)\sigma(b)-ab$$ $$s(a)=\sigma(a)-a$$ $$s(b)=\sigma(b)-b$$ $$s(a)s(b)-s(ab)=(\sigma(a)-a)(\sigma(b)-b)-(\sigma(a)\sigma(b)-ab)$$ $$=\sigma(a)\sigma(b)-a\sigma(b)-b\sigma(a)+ab-\sigma(a)\sigma(b)+ab$$ $$=2ab-a\sigma(b)-b\sigma(a)=(ab-b\sigma(a))+(ab-a\sigma(b))$$ $$=b(a-\sigma(a))+a(b-\sigma(b)) \leq 0,$$

where the last inequality holds since $x \leq \sigma(x)$ for all integers $x \geq 1$.

This completes the proof.

An Improvement to the Inequality When it is Known that $a > 1$ and $b > 1$

We have the penultimate equation and improved inequality:

$$s(a)s(b)-s(ab)=b(a-\sigma(a))+a(b-\sigma(b)) \leq -b - a = -(a+b),$$

where we have used the estimate $\sigma(x) \geq x + 1$, which holds for all integers $x>1$.

QUESTIONS

(1) Is the derivation of the improved inequality correct?

(2) Will it be possible to derive a better bound than the improved inequality? Or is this already best-possible?

Answer 1

(1) Looks good to me.

(2) The inequality cannot be simply improved since it is sharp when $\ a\ b\ $ are different primes.

However, inequality can be made more subtle and in this way -- indeed -- it can be improved. For instance (a not very systematic list of special cases follows):

• If relatively prime integers $\ a\ b\ $ are such $\ a>1\ $ and $\ b\ $ is composite then $\ \sigma(a)\ge a+1\ $ and $\ \sigma(b)\ge b+3$ hence

$$ \sigma(a\cdot b)\ \ge \ a\cdot b\ +\ (3\cdot(a+1)\ +\ b) $$

• If relatively prime integers $\ a\ b\ $ are both composite then $$ \sigma(a\cdot b)\ \ge\ (a+\sqrt a+1)\cdot(b+\sqrt b+1) $$

so that for $\ a<b\ $ we get

$$ \sigma(a\cdot b)\ \ge\ (a+3)\cdot(b+4)\,\ =\,\ a\cdot b \ +\ (3\cdot(a+b)\ +\ a\ +\ 12) $$

• If $\ a>1\ $ and $\ b:= p^x\cdot q^y\cdot t\ $ where $\ p\ q\ $ are different primes and $\ m:=\min(x\ y).\ $ Then the worst case is for $\ a:=5,\ p:=2,\ q:=3\ $ and $\ t:=1.\ $ Then

$$ \sigma(a\cdot b)\ \ge \sigma(5)\cdot\sigma(2^2)\cdot\sigma(3^2)\ = \ 6\cdot 7\cdot 13\ =\ 546 $$ or more generally we get the worst case when $\ p<q<a\ $ and $\ t:=1;\ $ then

$$ \sigma(a\cdot b)\ \ge\ (a+1)\cdot m^2\cdot(p^2+p+1)\cdot(q^2+q+1) $$

The obtained expression is so much larger than $\ a\cdot\ b\ $ that comparing them doesn't seem interesting (unless there are some special requirements).

Answer 2

(1) It looks correct to me.

(2) You cannot get a better inequality than $$s(a)s(b)-s(ab)\leq -(a+b)$$ since the equality of this inequality holds when $a,b$ are distinct primes.