<s>[INST] 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
 [/INST] 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:
$$
(x-1)(y-1)(z-1)\geq \frac{2x}{\sqrt{yz}}\frac{2y}{\sqrt{zx}}\frac{2z}{\sqrt{xy}}=8
$$
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