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The title says it all.

Here, $D(z) := 2z - \sigma(z)$ is the deficiency of $z \in \mathbb{N}$, and $\sigma(z)$ is the sum of the divisors of $z$.

Edited title on February 14 2017

ORIGINAL TITLE - Is $D(x) + D(y) \leq D(xy)$ true when $\gcd(x,y)=1$?

Added on February 14 2017

For an infinite family of counterexamples, consider $x=p=2$ and $y=q$ where $p$ and $q$ are distinct primes.

Then $$D(p) + D(q) = (p-1)+(q-1) = q \nleq D(pq) = 2pq - (p+1)(q+1) = (p-1)(q-1) - 2 = q - 3.$$

Revised Question

When does $D(x) + D(y) \leq D(xy)$ hold?

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    Always with $\gcd(x,y)=1$ and $xy$ deficient? – 2017-02-13
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    @Maczinga: With your clues, I have already been able to produce infinitely many counterexamples to my original question in the title. – 2017-02-13
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    My main interest would now be in my revised question. – 2017-02-13
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    To the question in the current title. No, there are infinitely many counterexamples. Take $(x,y)=(2,p)$ where $p\ge 5$ is an odd prime. Then, we get $\sigma(xy)=3p+3\lt 4p=2xy$ and $D(x)+D(y)=p\gt p-3=D(xy)$. (I'm writing this here as a comment because I don't know if you already know counterexamples.) – 2018-10-12
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    Yes, @mathlove. I am already aware of such counterexamples. (Please see details in my question above.) What I lacked was the extra condition $q \geq 5$. – 2018-10-12
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    @Jose Arnaldo Bebita Dris: Ah, OK. I have not read your question carefully. – 2018-10-13
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    "When does $D(x) + D(y) \leq D(xy)$ hold?" If you mean that you want to find a necessary and sufficient condition, then I think I cannot help you. I've just posted an answer showing a few necessary conditions. – 2018-10-14

3 Answers 3

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When does $D(x) + D(y) \leq D(xy)$ hold?

This answer shows a few necessary conditions.

Claim 1 : If $D(x) + D(y) \leq D(xy),x\gt 1,y\gt 1$ and $\gcd(x,y)=1$, then $x,y$ and $xy$ are deficient.

Proof : We have $$\begin{align}&D(x) + D(y) \leq D(xy) \\\\&\iff 2x-\sigma(x)+2y-\sigma(y)\le 2xy-\sigma(x)\sigma(y) \\\\&\iff \sigma(x)\sigma(y)-\sigma(x)-\sigma(y)\le 2xy-2x-2y \\\\&\iff \sigma(x)\sigma(y)-\sigma(x)-\sigma(y)+1\le 2xy-2x-2y+2-1 \\\\&\iff (\sigma(x)-1)(\sigma(y)-1)\le 2(x-1)(y-1)-1 \\\\&\iff \frac{\sigma(x)-1}{x-1}\cdot \frac{\sigma(y)-1}{y-1}\le 2-\frac{1}{(x-1)(y-1)}\ (\lt 2)\end{align}$$

So, since we have $$\frac{\sigma(x)-1}{x-1}\gt 1\qquad\text{and}\qquad \frac{\sigma(y)-1}{y-1}\gt 1$$ we have to have $$\frac{\sigma(x)-1}{x-1}\lt 2\qquad\text{and}\qquad \frac{\sigma(y)-1}{y-1}\lt 2,$$ i.e. $$\sigma(x)\lt 2x-1\lt 2x\qquad\text{and}\qquad \sigma(y)\lt 2y-1\lt 2x$$

Also, we have $$\begin{align}D(x)+D(y)\le D(xy)&\implies \sigma(xy)\le 2xy-2x-2y+\sigma(x)+\sigma(y) \\\\&\implies \sigma(xy)\lt 2xy-2x-2y+2x+2y \\\\&\implies \sigma(xy)\lt 2xy\qquad\square\end{align}$$

Claim 2 : There are infinitely many pairs $(x,y)$ such that $D(x) + D(y) \leq D(xy)$ and $\gcd(x,y)=1$.

Take $x=p,y=q$ where $p,q$ are two distinct odd primes.

$$D(x) + D(y) \leq D(xy)\iff (p-2)(q-2)\ge 3$$ which indeed holds.$\qquad\square$

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    Thank you for your answer, @mathlove. Unfortunately, I do not understand how you get the biconditional $$D(x) + D(y) \leq D(xy) \iff \frac{\sigma(x) - 1}{x - 1}\cdot\frac{\sigma(y) - 1}{y - 1} \leq 2 - \frac{1}{(x-1)(y-1)} < 2.$$ – 2018-10-14
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    @Jose Arnaldo Bebita Dris: I should have written the steps :) I've just added them. – 2018-10-14
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    My profuse thanks for the more comprehensive answer, @mathlove! Gladly accepting your answer now. =) – 2018-10-15
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Try the smallest non-trivial example of coprime integers: $$ D(2)+D(3)=1+2\color{red}{\not\le} 0=D(6).$$

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    Oups! We wrote pratically the same! – 2017-02-13
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    Kindly check the updated title - my apologies for missing out on the *deficient* $xy$ earlier... – 2017-02-13
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Take $2^n$ and $3^n$. Then the inequality do not hold.

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    Please check the updated title. – 2017-02-13
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    I think that if you take $x=p$ and $y=2p$ ($p$ prime >2) then the inequality holds. – 2017-02-13
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    Indeed, if $x=p$ and $y=2p$ (where $p$ is an odd prime), then $$D(x) + D(y) = (p - 1) + (p - 3) = 2(p - 2)$$ while $$D(xy) = D(2p^2) = 4p^2 - \sigma(2p^2) = 4p^2 - 3(p^2 + p + 1) = p^2 - 3p - 3.$$ It remains to verify $2p - 4 \leq p^2 - 3p - 3$. This implies $p^2 - 5p + 1 \geq 0$. This is not true for $p=3$, but holds for $p \geq 5$. – 2017-02-14