GCD and linear combinations

Counterexample to the first statement. $a=b=1$: $\gcd(1,1)=1$, $\operatorname{lcm}(1,1)=1$, $\gcd(2,1)=1$. Actually this works! Try $a=b=2$: $\gcd(2,2)=2$, $\operatorname{lcm}=2$, $\gcd(4,2)=2$. Still works. Hmm, maybe the statement is true for $a=b$. Try $a=2$, $b=3$: $\gcd=1$, $\operatorname{lcm}=6$, $\gcd(5,6)=1$. Still works. The statement is actually always true: $\gcd(a+b,\operatorname{lcm}(a,b))=\gcd(a,b)$ (as shown in an earlier exercise). So the first statement "is false" is incorrect; it is true. Let me instead show a counterexample to a different statement, e.g. $\gcd(a+b,ab)=\gcd(a,b)$.

Counterexample to $\gcd(a+b,ab)=\gcd(a,b)$ in general. $a=2$, $b=4$: $\gcd(2,4)=2$, $\gcd(6,8)=2$. Still works! Try $a=2$, $b=2$: $\gcd(2,2)=2$, $\gcd(4,4)=4\neq2$. Counterexample: $a=b=2$.

Second part. If $\gcd(a,b)=1$, then every prime dividing $a+b$ and $ab$ would divide $a$ or $b$ (by Euclid's lemma), and then also the other, contradicting coprimality. So $\gcd(a+b,ab)=1=\gcd(a,b)$.

Conclusion. The general identity $\gcd(a+b,ab)=\gcd(a,b)$ fails (e.g. $a=b=2$), but holds when $\gcd(a,b)=1$.