AMC 12 Mock Exam 3 Problem 20

Fix $c$. Then $a+b=6-c$ and $ab=9-c(a+b)=9-c(6-c)=(c-3)^2$, so $abc=c(c-3)^2$. Real $a,b$ exist iff $(a+b)^2\ge4ab$, i.e., $(6-c)^2\ge4(c-3)^2$, which simplifies to $3c^2-12c\le0$, i.e., $0\le c\le4$.

For $0\le c\le4$, $$4-c(c-3)^2=-c^3+6c^2-9c+4=(c-1)^2(4-c)\ge0,$$ so $abc\le4$, with equality at $c=4$ (then $a=b=1$; check: $1+1+4=6$, $1+4+4=9$). The maximum is $\boxed{4}$.