An inequality relating the deficiency and sum of aliquot divisors

Question

In what follows, we let $n$ be a positive integer. The classical sum of divisors of $n$ is given by $\sigma_1(n)$.

Denote the deficiency of $n$ by $D(n)=2n-\sigma_1(n)$, and denote the sum of aliquot divisors of $n$ by $s(n)=\sigma_1(n)-n$.

(Hereinafter, I shall abbreviate the notation for the divisor sum $\sigma_1$ simply as $\sigma$.)

Here is my question:

Is the following inequality true in general, where $\gcd(a,b)=1$? $$D(ab) < D(a)s(b)$$

MY ATTEMPT

Since $\gcd(a,b)=1$, we can rewrite $$D(ab) = 2ab - \sigma(ab) = 2ab - \sigma(a)\sigma(b)$$ $$D(a)s(b) = (2a - \sigma(a))(\sigma(b) - b) = -2ab - \sigma(a)\sigma(b) + 2a\sigma(b) + b\sigma(a).$$

Therefore, we get $$D(ab) - D(a)s(b) = \bigg(2ab - \sigma(a)\sigma(b)\bigg) - \bigg(-2ab - \sigma(a)\sigma(b) + 2a\sigma(b) + b\sigma(a)\bigg),$$ from which we obtain $$D(ab) - D(a)s(b) = 4ab - 2a\sigma(b) - b\sigma(a) = ab + (2ab - 2a\sigma(b)) + (ab - b\sigma(a)) = ab + 2a(b - \sigma(b)) + b(a - \sigma(a)) = ab - 2as(b) - bs(a).$$ Alas, this is where I get stuck. I currently do not see an easy way to get an upper bound of $0$ for $$ab - 2as(b) - bs(a)$$ when $\gcd(a,b)=1$.

Answer 1

No, it isn't.

There are infinitely many counterexamples.

If $a,b$ are distinct odd prime numbers, then we have $\gcd(a,b)=1$ and $$\begin{align}D(ab)-D(a)s(b)&=(2ab-(a+1)(b+1))-(2a-a-1)(b+1-b) \\\\&=ab-2a-b \\\\&=(a-1)(b-2)-2 \\\\&\gt 0\end{align}$$

Answer 2

I tried to check for counterexamples to the inequality $$D(xy) < D(x)s(y)$$ where $\gcd(x,y)=1$, using the following Pari-GP script in Sage Cell Server:

for(x=1, 1000, for(y=1, 1000, if(2*x*y-sigma(x*y) >= (2*x-sigma(x))*(sigma(y)-y) && gcd(x,y)==1,print(x,"   ",y))))

There were several lines returned. This shows that the inequality is not true in general.