1. **State the problem:** We want to prove that if $\gcd(a,b) = 1$, then $\gcd(a+b, ab) = 1$. 2. **Recall the definition:** $\gcd(x,y)$ is the greatest positive integer that divides both $x$ and $y$. 3. **Assume for contradiction:** Suppose $d = \gcd(a+b, ab)$ and $d > 1$. 4. Since $d$ divides $a+b$ and $ab$, it must divide any integer linear combination of these two numbers. 5. Consider the combination $d \mid (a+b)$ and $d \mid ab$. 6. Then $d$ divides $a(a+b) - ab = a^2$. 7. Similarly, $d$ divides $b(a+b) - ab = b^2$. 8. So $d$ divides both $a^2$ and $b^2$. 9. Since $d$ divides $a^2$ and $b^2$, it must divide $a$ and $b$ (because any prime factor dividing $a^2$ divides $a$). 10. But this contradicts the assumption that $\gcd(a,b) = 1$. 11. Therefore, our assumption that $d > 1$ is false, so $\gcd(a+b, ab) = 1$. **Final answer:** $\boxed{\gcd(a+b, ab) = 1}$ if $\gcd(a,b) = 1$.