Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$

Question

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

Answer 1

We have $x\left(\frac1x+\frac1y+\frac1z\right)=x\implies x-1=\frac xy+\frac xz\overset{AM-GM}{≥}\frac{2x}{\sqrt{yz}}$ and similarly for $y$ and $z$. So we have: $$ (x-1)(y-1)(z-1)\geq \frac{2x}{\sqrt{yz}}\frac{2y}{\sqrt{zx}}\frac{2z}{\sqrt{xy}}=8 $$

Answer 2

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

Now $$(x-1)(y-1)(z-1)=(xy-x-y+1)(z-1)=$$ $$=(xyz-xz-yz-xy)+(x+y+z)-1=0+(x+y+z)-1\ge 9-1=8$$