Tricky inequality involving 3 variables Let $x, y$ and $z$ be three real numbers satisfying the following conditions:
$$0 < x \leq y \leq z$$ 
AND
$$xy + yz + zx = 3$$
Prove that the maximum value of $(x y^3 z^2)$ is $2.$
I tried using the weighted AM-GM inequality, but to no avail as the powers 1,2 and 3 are giving me a hard time. How should I proceed? Thanks in advance.

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!
