Let $x=\frac{a}{2\sqrt2}$, $y=\sqrt2b$ and $z=\sqrt2c$.

Hence, $c\geq b$ and by AM-GM: $$6=4bc+ab+ac\geq6\sqrt[6]{(bc)^4(ab)(ac)}=6\sqrt[6]{a^2b^5c^5}\geq6\sqrt{a^2b^6c^4},$$ which gives $$1\geq ab^3c^2=\frac{1}{2}xy^3z^2.$$ The equality occurs for $x=\frac{1}{2\sqrt2}$ and $y=z=\sqrt2$

and we are done!