If $(a,b)=1$ then prove $(a+b, ab)=1$. 
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.

$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?

Here is another proof using Bezout's Identity:
$$
\begin{align}
ma+nb&=1\tag{1}\\
(n-m)b&=1-m(a+b)\tag{2}\\
(m-n)a&=1-n(a+b)\tag{3}\\
-(m-n)^2ab&=1-(m+n)(a+b)+mn(a+b)^2\tag{4}\\
1&=((m+n)-mn(a+b))\color{#C00000}{(a+b)}-(m-n)^2\color{#C00000}{ab}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $(a,b)=1$ and Bezout's Identity
$(2)$: subtract $m(a+b)$ from both sides of $(1)$
$(3)$: subtract $n(a+b)$ from both sides of $(1)$
$(4)$: multiply $(2)$ and $(3)$
$(5)$: rearrange and collect terms of $a+b$ and $ab$ from $(4)$
Bezout's Identity and $(5)$ say that $(a+b,ab)=1$.

Henning Makholm and Bill Dubuque also give Bezout Identity based proofs. The coefficient of $ab$ is the same in all of our answers, but the coefficient of $a+b$ in mine looks different from theirs. They are the same however:
$$
\begin{align}
(m+n)-mn(a+b)
&=(m+n)\overbrace{(ma+nb)}^{1}-mn(a+b)\\
&=m^2a+n^2b+mn(a+b)-mn(a+b)\\
&=m^2a+n^2b
\end{align}
$$
