Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$ I need to prove that: 

If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$

So far I used what's given so I have:

$ax+by=1$ (I wrote the gcd of a and b as a linear combination)

and 

$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)

where do I go from here?
Below is a proof that is messier but more primitive than Yiorgos' proof. By Bezout's Identity, we have that there exists $x,y \in \mathbb{Z}$ such that

$$ax + by = 1.$$

Squaring both sides, we see that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = 1.$$

But notice that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2).$$

And hence, those same $x,y$ as above give

$$ ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2) = 1.$$

So it must be that $\gcd{(a+b,ab)} = 1$.
Proof broken into lemmas:

Lemma 1:  If $n$ is rel. prime to $a$ and $n$ is rel. prime to $b$, then $n$ is rel. prime to $ab$.

Proof:  multiply together the Bezout identities for the two hypotheses.

Lemma 2:  If $a$ and $b$ are rel. prime, then both are rel. prime to $a+b$.

Proof:  Its enough to prove if  $a$ and $b$ are rel. prime then $b$ is rel. prime to $a+b$.
We have the Bezout identity:  $a x + by = 1$.  Now add and subtract $bx$ on the left.

Conclusion:  We are given $a$ and $b$ rel. prime.  By the 2nd lemma, $a+b$ is rel. prime to both $a$ and $b$.  By the first lemma, $a+b$ is rel. prime to $ab$.