Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$.

$$(1+x)(1+y)(1+z) \ge 8(1-x)(1-y)(1-z) \Leftrightarrow $$ $$(2x+y+z)(x+2y+z)(x+y+2z) \ge 8(y+z)(x+z)(x+y)$$

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 $$2\sqrt{(1-x)(1-z)}\le 1+y \,.$$ $$2\sqrt{(1-x)(1-y)}\le 1+z \,.$$

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 $$(1+x)(1+y)\ge2(x+y)$$