Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)} $

Question

Let $x,y,z$ be real numbers in the range of $(0,1)$. Prove that $$\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}.$$

Answer 1

\begin{eqnarray} &&[\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)}][xyz+(1-x)(1-y)(1-z)]\\ &=&[\frac{yz}{1-y}+\frac{(1-x)(1-z)}{x}]+[\frac{xz}{1-z}+\frac{(1-x)(1-y)}{y}]+[\frac{xy}{1-x}+\frac{(1-y)(1-z)}{z}]\\ &\ge&[\frac{z}{1-y}+\frac{1-z}{x}-1]+[\frac{x}{1-z}+\frac{1-x}{y}-1]+[\frac{y}{1-x}+\frac{1-y}{z}-1]\\ &\geq& 6\sqrt[6]{\frac{z}{1-y}\cdot\frac{1-z}{x}\cdot\frac{x}{1-z}\cdot\frac{1-x}{y}\cdot\frac{y}{1-x}\cdot\frac{1-y}{z}}-3=3 \end{eqnarray}

Answer 2

It's obviously true after following substitution and full expanding.

$x=\frac{a}{a+1}$, $y=\frac{b}{b+1}$ and $z=\frac{c}{c+1}$, where $a$, $b$ and $c$ are positives.