<s>[INST] Struggle with finding equality condition in an AM-GM inquality problem Given $0<x,y,z<1$. Prove that
$$\dfrac{1-x}{1+y+z}+\dfrac{1-y}{1+z+x}+\dfrac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z).$$
I've thought that the equality holds when $x=y=z$ and I'm thinking ways to use AM-GM inequality. However, it turns out that
$$3\dfrac{1-x}{1+2x}=3(1-x)^3 \Longleftrightarrow (1-2x+x^2)(1+2x)=1\Longleftrightarrow 2x^3-3x^2=0\Longleftrightarrow x=\dfrac{3}{2},$$
which is totally wrong due to $0<x,y,z<1$! Thank you a lot!
 [/INST] AM-GM seems like an unlikely choice in this problem.  I am presenting a different approach.  
Note that
\begin{align}\frac{1-x}{1+y+z}-(1-x)(1-y)(1-z)&=\frac{1-x}{1+y+z}\big(1-(1-y-z+yz)(1+y+z)\big)
\\&=\frac{1-x}{1+y+z}\big(y^2(1-z)+z^2(1-y)+yz\big)\geq 0\end{align}
for all $x,y,z\in[0,1]$.  Similarly,
\begin{align}\frac{1-y}{1+z+x}-(1-x)(1-y)(1-z)\geq 0\end{align}
and
\begin{align}\frac{1-z}{1+x+y}-(1-x)(1-y)(1-z)\geq 0.\end{align}
Summing all three inequalities above yields 
$$\frac{1-x}{1+y+z}+\frac{1-y}{1+z+x}+\frac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z)$$
for all $x,y,z\in[0,1]$.  There are only two equality cases, i.e., $x=y=z=0$ and $x=y=z=1$.
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