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.

Let $$P(x;y;z)=\frac{8}{27}(x+y+z)^3-(x+y)(y+z)(x+z)$$
$P(1;1;1)\ge0; P(1;1;0)\ge0$ and $P(1;0;0)\ge0$
Then $$P(x;y;z)\ge0$$
Hence,
$$(x+y)(y+z)(x+z) ≤ \frac{8}{27}(x+y+z)^3$$
