Assuming $a, b, c > 1$, then $(a, bc) = 1$ implies $b \not\mid a$, for if $b \mid a$, then we'd have $1 = (a, bc) \geq b > 1$ which is a contradiction. Likewise, $(a, bc) = 1$ also implies $c \not\mid a$.
So $b > 1$ and $(a, bc) = 1$ implies $b \not\mid a$ which implies $(a, b) < b > 1$. Can it be shown also that $(a, b) = 1$?
If $d \mid a, b$, then $d \mid a, bc$. The contrapositive is "if $d \not\mid a, bc$, then $d \not\mid a, b$. From $(a, bc) = 1$, we know $d > 1 \not\mid a, bc$, therefore $d > 1 \not\mid a, b$, therefore $(a, b) = 1$.
– joseville May 07 '22 at 20:35