Can You Prove the GCD Inequality for Natural Numbers?

[lfdahl]

Prove, that for all natural numbers, $a$ and $b$, with $b > a$:

\[\frac{ab}{(a,b)}+\frac{(a+1)(b+1)}{(a+1,b+1)}\geq \frac{2ab}{\sqrt{b-a}}\]

where $(m,n)$ denotes the greatest common divisor of the natural numbers $m$ and $n$.

 

[Albert1]

Prove, that for all natural numbers, $a$ and $b$, with $b > a$:

\[\frac{ab}{(a,b)}+\frac{(a+1)(b+1)}{(a+1,b+1)}\geq \frac{2ab}{\sqrt{b-a}}\]

where $(m,n)$ denotes the greatest common divisor of the natural numbers $m$ and $n$.

Spoiler

let $A=\dfrac{ab}{(a,b)}+\dfrac{(a+1)(b+1)}{(a+1,b+1)}$
$B=\dfrac{2ab}{\sqrt{b-a}}$
using $a\times b=(a,b)\times [a,b]$
where $(a,b)$ denotes the greatest common divisor of the natural numbers $a$ and $b$
and $[a,b]$ denotes the least common multiple of the natural numbers $a$ and $b$
we have $min(A)=b+b+1\leq A=[a,b]+[a+1,b+1]\leq ab+(a+1)(b+1)$
$=2ab+a+b+1=max(A)$
$B\leq 2ab=max(B)$
if we can find a solution $min(A)\geq max(B)$
that is to find solution $2b+1\geq 2ab$ then we have the proof
in this case $a=1,b>a$
in general we may set $b-a=k>0$
to find $\sqrt k(2b+1)\geq 2ab$
we get $\sqrt k=a,or \,\, k=a^2$

 

[lfdahl]

Spoiler

let $A=\dfrac{ab}{(a,b)}+\dfrac{(a+1)(b+1)}{(a+1,b+1)}$
$B=\dfrac{2ab}{\sqrt{b-a}}$
using $a\times b=(a,b)\times [a,b]$
where $(a,b)$ denotes the greatest common divisor of the natural numbers $a$ and $b$
and $[a,b]$ denotes the least common multiple of the natural numbers $a$ and $b$
we have $min(A)=b+b+1\leq A=[a,b]+[a+1,b+1]\leq ab+(a+1)(b+1)$
$=2ab+a+b+1=max(A)$
$B\leq 2ab=max(B)$
if we can find a solution $min(A)\geq max(B)$
that is to find solution $2b+1\geq 2ab$ then we have the proof
in this case $a=1,b>a$
in general we may set $b-a=k>0$
to find $\sqrt k(2b+1)\geq 2ab$
we get $\sqrt k=a,or \,\, k=a^2$

Thankyou, Albert for your approach to the problem. Well done.

Here´s another approach:

Spoiler

By applying the AM-GM inequality we get:
\[\frac{ab}{(a,b)}+\frac{(a+1)(b+1)}{(a+1,b+1)}\geq 2 \sqrt{\frac{a(a+1)b(b+1)}{(a,b)(a+1,b+1)}} > \frac{2ab}{\sqrt{(a,b)(a+1,b+1)}}\]
In order to complete the solution, we now show that:
\[b-a \ge (a,b)(a+1,b+1)\]
Indeed, $(a,b)$ and $(a+1,b+1)$ both divide $b-a$, since the greatest common divisor of two
numbers also divides their difference. On the other hand, $(a,b)$ and $(a+1,b+1)$ are coprimes.
Therefore the product $(a,b)(a+1,b+1)$ divides $b-a$. Hence the inequality is proved.