Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ How can I prove that if $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1,$$ then $(x-1)(y-1)(z-1)  \geq 8$? 
Edit: 
$x,y,z \in \mathbb R_{>0} $
Thanks

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
$$xyz=xy+yz+zx$$
$$AM \ge GM \Rightarrow (x+y+z)\left (\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\ge 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac 1{xyz}}=9 \Rightarrow$$
$$\Rightarrow (x+y+z)\cdot 1 \ge 9$$
Now $$(x-1)(y-1)(z-1)=(xy-x-y+1)(z-1)=$$
$$=(xyz-xz-yz-xy)+(x+y+z)-1=0+(x+y+z)-1\ge 9-1=8$$
