Prove $(x+y)(y+z)(x+z)≤\frac8{27}(x+y+z)^3$ The entire exercise was like
$$8xyz≤(x+y)(y+z)(x+z)≤\frac8{27}(x+y+z)^3\quad x,y,z>0$$ 
and I managed to solve (demonstrate) the first inequality (from the left) using
$$\frac{x+y}2≥\sqrt{xy}$$
but I can't solve the second one. I tried several ideas but I haven't succeeded.
Thanks a lot.

Hint: The arithmetic mean of $\frac{x+y}2$, $\frac{x+z}2$, $\frac{y+z}2$ is $\frac{x+y+z}{3}$, whereas the geometric mean is $\sqrt[3]{\frac{(x+y)(y+z)(x+z)}{8}}$
