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.

$$(1+x)(1+y)(1+z) \ge 8(1-x)(1-y)(1-z) \Leftrightarrow $$
$$(2x+y+z)(x+2y+z)(x+y+2z) \ge 8(y+z)(x+z)(x+y)$$
Let $a=x+y, b=x+z, c=y+z$. Then the inequality to prove is
$$(a+b)(a+c)(b+c) \ge 8abc \,,$$
Which follows immediately from AM-GM:
$$a+b \ge 2 \sqrt{ab}$$
$$a+c \ge 2 \sqrt{ac}$$
$$b+c \ge 2 \sqrt{bc}$$
Simplification The solution above can be simplified the following way:
By AM-GM
$$2\sqrt{(1-y)(1-z)}\le 1-y+1-z=1+x \,.$$
Similarly
$$2\sqrt{(1-x)(1-z)}\le 1+y \,.$$
$$2\sqrt{(1-x)(1-y)}\le 1+z \,.$$
