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?

From $xyz=(1-x)(1-y)(1-z)$ we get
$$1-2xyz=\sum_{cyc}x(1-y)\geq 3\sqrt[3]{xyz(1-x)(1-y)(1-z)}=3\sqrt[3]{x^2y^2z^2},$$
Which, on setting $xyz=\dfrac {t^3}{8},$ gives us $4\ge t^3+3t^2$. This factors into
$$0\ge (t-1)(t+2)^2,$$
Hence $t\le 1$ and $xyz\le \dfrac 18$. So,
$$\sum_{cyc}x(1-y)=1-2xyz\ge\frac34.$$
Equality occurs at $x=y=z=\dfrac 12.$
