Conjecture that $ \frac{\gcd(a+b,ab)}{\gcd(a,b)} \mid \gcd(a,b)$ I have discovered some exercise type conjectures which I can't prove and this is one of them:

Given positive integers $a,b$, then
  $$ \frac{\gcd(a+b,ab)}{\gcd(a,b)}\ \bigg|\ \gcd(a,b)$$

Can this be proved or disproved?

From time to time, when testing my growing math packages BigZ and Forthmath, I recognize some patterns which I can't prove or disprove (or even have the ambition to). I post them here with the hope that it will not annoy too much. I hope you can bear with it.
 A: Write $\gcd(a,b)=d$, then $a=da',b=db'$ and thus $\frac{\gcd(a+b,ab)}{d}=\gcd(a'+b',a'b'd)$ where $\gcd(a',b')=1$. Notice now that $\gcd(a'+b',a'b')=1$ since $a'(a'+b')-a'b'={a'}^2$ and thus $\gcd(a'+b',a'b')|{a'}^2 \implies \gcd(a'+b',a'b')|a'$ and $\gcd(a'+b',a'b')|a'+b' \implies \gcd(a'+b',a'b')|a',b' \implies \gcd(a'+b',a'b')=1$. This means that $\gcd(a'+b',a'b'd)=\gcd(a'+b',d)$ and thus it divides $d$ by definition. So your conjecture is indeed true.
A: Taking $D=(a,b)$, then $a=DA$ and $b=DB$, $(ab,a+b)=D(A+B,ABD)$ and $(A,B)=1$. So you want to know if $(ab,a+b)/D$ divides $D$, i.e. $(A+B,ABD)|D$? well, let's see if the prime common divisors between $A+B$ and $ABD$ are divisor of $D$ as well.
If $p$ is a prime common divisor of $A+B$ and $ABD$, by Gauss lemma, $p|A$ or $p|B$ or $p|D$. If $p|A$ or $p|B$ there is a contradiction with the coprimality of $A$ and $B$ $p$ cause if $p|A$ then $p|(A+B)-A=B$. So $p|D$. 
A: https://en.wikipedia.org/wiki/P-adic_order
Let $h = \gcd(a+b, ab)$ and $g = \gcd(a,b).$
For each prime factor $p$ including $2,$ two cases:
(I) $$ a = p^k u, b = p^j v $$
   with $k > j$ and $u,v \neq 0 \pmod p.$
Then $p$-adic valuation $\nu_p(g) = j.$ Next $\nu_p(ab) = k+j$ while $\nu_p(a+b) = j.$  Put together, $\nu_p(g) = \nu_p(h).$
(II) $$ a = p^k u, b = p^k v $$
   with  $u,v \neq 0 \pmod p.$
Then $p$-adic valuation $\nu_p(g) = k.$ Next $\nu_p(ab) = 2k$ while $\nu_p(a+b) \geq  k.$ Then
$$ k \leq \nu_p(h) \leq 2k $$
Put together, $\nu_p(h) \leq 2\nu_p(g).$
In either case, combining all primes, 
$$  h | g^2 $$
Oh, note that we do have $g | h$ and can write $$ \frac{h}{g} \; | \; g  $$
A: Recall $(a,b) = (a\!+\!b,{\rm lcm}(a,b))\,$ so $\,(a\!+\!b,ab)\mid (a,b)^2\! = (a\!+\!b,{\rm lcm}(a,b))^2$ by $\,ab\mid {\rm lcm}(a,b)^2$
A: Let $m=\gcd(a,b)$. Now $m\mid a$ and $m\mid b$, so $m^2\mid ab$. 
Let $a=a_1m$, $b=b_1m$, for some positive integers $a_1$, $b_1\ge1$, then $a+b=ma_1+mb_1=m(a_1+b_1)$, so $m\mid a+b$. 
There are three cases to consider:
Case 1:
If $m\nmid(a_1+b_1)$ then $\gcd(a+b,ab)=m$, and we have
$$\frac{\gcd(a+b,ab)}{\gcd(a,b)}=\frac{m}{m}=1\mid\gcd(a,b)=m$$
and the conjecture is true in this case. 
Now we need to check whether $m^2\mid a+b=m(a_1+b_1)$, which is equivalent to $m\mid(a_1+b_1)$. 
Case 2:
If $a_1+b_1=mc$, for some integer $c\ge1$, then $m\mid(a_1+b_1)$, and so $m^2\mid a+b=m(a_1+b_1)$. In this case $\gcd(a+b,ab)=m^2$ (note we cannot have $\gcd(a+b,ab)>m^2$ as $\gcd(a,b)=m$, therefore $m^3\nmid ab$), and we have
$$\frac{\gcd(a+b,ab)}{\gcd(a,b)}=\frac{m^2}{m}=m\mid\gcd(a,b)=m$$
and the conjecture is true in this case.
Consider lastly the case:
Case 3:
Let us assume
$$m<\gcd(a+b,ab)=\gcd(m(a_1+b_1),m^2a_1b_1)=m\gcd(a_1+b_1,ma_1b_1)<m^2$$
which simplifies to
$$1<\gcd(a_1+b_1,ma_1b_1)<m$$
We cannot have $m\mid(a_1+b_1)$, as then $\gcd(a_1+b_1,ma_1b_1)=m$, a contradiction. Further simplification gives
$$1<\gcd(a_1+b_1,a_1b_1)<m$$
Let $p$ be s.t. $1<p<m$ and $\gcd(a_1+b_1,a_1b_1)=p$. Then $p\mid a_1+b_1$ and $p\mid a_1b_1$, then $p\mid a_1$ or $b_1$. If $p\mid a_1$ then $p\mid a_1+b_1$ implies $p\mid(a_1+b_1)-a_1=b_1$, a contradiction since $\gcd(a_1,b_1)=1$. 
So $\gcd(a_1+b_1,a_1b_1)=1$, and it follows $\gcd(a+b,ab)$ can take no value in $(m,m^2)$.
The conjecture is true.
