Find the minimum possible value of $x(1-z)+y(1-x)+z(1-y)$ It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$
Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$
Using the AM-GM inequality concepts, I can write that the value is minimum when 
$$x(1-z)=y(1-x)=z(1-y)$$
What else can I conclude from the given information?

Making the substitutions $\displaystyle \frac{1-x}{x} = a, \frac{1-y}{y} = b$ and $\displaystyle \frac{1-z}{z} = c$,
Then, $abc = 1$ and 
$\begin{align}\displaystyle \sum\limits_{cyc} x(1-z) &= \sum\limits_{cyc} \frac{c}{(1+c)(1+a)} \\&= \frac{\sum\limits_{cyc} c(1+b)}{(1+a)(1+b)(1+c)} \\ &= \frac{\sum\limits_{cyc} c(1+b)}{1+\sum\limits_{cyc} a + \sum_{cyc} ab + abc} = \frac{\sum\limits_{cyc} a + \sum_{cyc} ab}{2+\sum\limits_{cyc} a + \sum_{cyc} ab}\end{align}$
Assume the minimum value of $\sum\limits_{cyc} x(1-z)$ is $m$ (a minumum exists since $\sum\limits_{cyc} x(1-z) > 0$).
Then, $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge m\left(2+\sum\limits_{cyc}a + \sum\limits_{cyc} ab\right)$
i.e. $\displaystyle \sum\limits_{cyc} a + \sum\limits_{cyc} ab \ge \frac{2m}{1-m}$.
But, we know $\displaystyle \sum\limits_{cyc} a \ge 3\sqrt[3]{abc} = 3$ and $\displaystyle \sum\limits_{cyc} ab \ge 3\sqrt[3]{a^2b^2c^2} = 3$ (from Am-Gm inequality and $abc = 1$)
Thus, $\displaystyle \frac{2m}{1-m} = 6 \implies m = \frac{3}{4}$.
2nd approach: 
$\displaystyle xyz = (1-x)(1-y)(1-z) \implies \sum\limits_{cyc} x(1-z) = \sum\limits_{cyc} x - \sum\limits_{cyc} xy = 1-2xyz$
and, $\displaystyle xyz = \frac{1}{\prod\limits_{cyc} \left(1+\frac{1-x}{x}\right)} \le \frac{1}{\prod\limits_{cyc} 2\sqrt{\frac{1-x}{x}}} = \frac{1}{8}$ (by Am-Gm inequality)
Thus, $\displaystyle \sum\limits_{cyc} x(1-z) \ge \frac{3}{4}$
