Struggle with finding equality condition in an AM-GM inquality problem

Question

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!

Answer 1

It should be $$3x^2-2x^3\geq0,$$ which is true for $0<x<1.$

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

Thus, we need to prove that $$\sum_{cyc}\frac{(a+1)^2(b+1)^2}{3ab+2a+2b+1}\geq3.$$ Now, by C-S $$\sum_{cyc}\frac{(a+1)^2(b+1)^2}{3ab+2a+2b+1}\geq\frac{\left(\sum\limits_{cyc}(ab+2a+1)\right)^2}{\sum\limits_{cyc}(3ab+4a+1)}.$$ Thus, it's enough to prove that $$\left(\sum\limits_{cyc}(ab+2a+1)\right)^2\geq3\sum\limits_{cyc}(3ab+4a+1)$$ or $$\sum_{cyc}(a^2b^2+2a^2bc+4a^2+5ab+4a^2b+4a^2c+4abc)\geq0,$$ which is obvious.

We can use also another C-S.

Indeed, we need to prove that: $$\sum_{cyc}\frac{1}{(1-x)(1-y)(1+x+y)}\geq3.$$ Now, $$\sum_{cyc}\frac{1}{(1-x)(1-y)(1+x+y)}\geq\frac{9}{\sum\limits_{cyc}(1+x+y)(1-x)(1-y)}.$$ Id est, it's enough to prove that $$3\geq\sum_{cyc}(1+x+y)(1-x-y+xy)$$ or $$\sum_{cyc}(2x^2+xy)\geq\sum_{cyc}(x^2y+x^2z)$$ or $$\sum_{cyc}(x^2(1-y)+x^2(1-z)+xy)\geq0,$$ which is obvious.

Answer 2

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 $x=y=z=1$.