Show that if $a$ and $b$ are positive integers then $(a, b) = (a + b, [a, b])$. Show that if $a$ and $b$ are positive integers then $(a, b)=(a + b, [a, b])$.
I was thinking that since $[a, b]=LCM(a, b)=\frac{ab}{(a, b)}$ that if $d= (a + b, [a, b])$, then $d|[a,b]$ and thus $d|(a, b)$ since $(a, b)|[a, b]$
Then I would just have to prove that $(a+b, (a, b)) = (a, b)$.
Is that the right way to go?

Another  way :  Let $(a,b)=d$ and $\displaystyle \frac aA=\frac bB=d\implies (A,B)=1$
So, $\displaystyle(a+b, [a,b])=(d(A+B), dAB)=d(A+B,AB)$
Now, if $D$ divides both $ A+B, AB; D$ will divide $A(A+B)-AB=A^2$ and $D$ will divide $B(A+B)-AB=B^2$
$\displaystyle\implies D$ will divide $(A^2,B^2)=(A,B)^2=1\implies D=1$
