Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$. 
If $x>0,y>0,z>0$ and $x+y+z=1$, prove that $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z).$$

Trial: Here  $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z) \\ \implies  (1+x)(1+y)(1+z)\ge 8(y+z)(x+z)(x+y)$$ I am unable to solve the problem. Please help.

since 
$$t=(x+y)(y+z)(x+z)\le\left(\dfrac{2(x+y+z)}{3}\right)^3=\dfrac{8}{27}$$
and we
have
$$(1-x)(1-y)\ge 0$$
then
$$(1+x)(1+y)\ge2(x+y)$$
