AM-GM Application

$\begin{aligned} \frac 1x + \frac 1y + \frac 1z & = 1 \\ \frac {xy+yz+zx}{xyz} & = 1 \\ \implies \color{#3D99F6} xy+yz+zx & \color{#3D99F6}= xyz \end{aligned}$

Now, we have:

$\begin{aligned} (x-1)(y-1)(z-1) & = {\color{#3D99F6}xyz - xy-yz-zx} + x + y + z - 1 \\ & = {\color{#3D99F6}0}+ {\color{#D61F06}x + y + z} -1 & \small {\color{#D61F06} \text{By AM-GM inequality}} \\ & \ge {\color{#D61F06}3\sqrt[3]{xyz}} - 1 & \small {\color{#D61F06} \text{Equality occurs when }x=y=z=3} \\ & \ge {\color{#D61F06}9} - 1 = \boxed{8} \end{aligned}$

Reference: AM-GM inequality

Expanding $(x-1)(y-1)(z-1)$ we obtain $xyz-xy-xz-yz+x+y+z-1$ , and because of the equality stated in the problem, we know $xy+xz+yz=xyz$ . Thus, the problem now is to find the minimal value of $x+y+z-1$ . By Cauchy-Schwarz Inequality(sorry for not using AM-GM Inequality), $(\sqrt{x}^2+\sqrt{y}^2+ \sqrt{z} ^2)(1/(\sqrt{x}^2)+1/(\sqrt{y}^2)+1/(\sqrt{z}^2))>=3^2$ <=> $x+y+z-1>=8$ .

xyz=x+y+z , $\frac{1}{x}$ $+$ $\frac{1}{y}$ $+$ $\frac{1}{z}$ >= $\frac{(1+1+1)^2}{x+y+z}$ , so x+y+z>=9 --> (x-1)(y-1)(z-1)=x+y+z-1>=8

1/x +1/y +1/z =1 yz+xz+xy=xyz yz+xz=xyz-xy ->z(x+y)=xy(z-1) eq1 similarly, y(x+z)=xz(y-1) eq2 and x(z+y)=zy(x-1) eq3 ultiplying eq 1,2, and 3 z^2x^2y^2(x-1)(z-1)(y-1)=xyz(x+y)(z+y)(x+z) (x-1)(z-1)(y-1)=[(x+y)(z+y)(x+z)]/xyz so min(x-1)(z-1)(y-1)=min[(x+y)(z+y)(x+z)]/xyz [(x+y)(z+y)(x+z)]/xyz=(1+y/x)(1+z/y)(1+x/z) =1+z/y+y/x+z/x+x/z+x/y+y/z+1 so, min(x-1)(z-1)(y-1)=2+min z/y+y/x+z/x+x/z+x/y+y/z using am-gm z/y+y/x+z/x+x/z+x/y+y/z >=6(6throot(z/y)(y/x)(z/x)(x/z)(x/y)(y/z)) z/y+y/x+z/x+x/z+x/y+y/z >=6 so min(x-1)(z-1)(y-1)=2+6=8