\(S>\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\Rightarrow S>1\)

\(S< \frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}\Rightarrow S< 2\)

\(\Rightarrow1< S< 2\)

\(xy+yz+zx-xyz=1-x-y-z+xy+yz+zx-xyz\)

\(=\left(1-x\right)-y\left(1-x\right)-z\left(1-x\right)+yz\left(1-x\right)\)

\(=\left(1-x\right)\left(1-y-z+yz\right)=\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

\(xy+yz+zx+xyz+2=1+x+y+z+xy+yz+zx+xyz\)

\(=\left(1+x\right)+y\left(1+x\right)+z\left(1+x\right)+yz\left(1+x\right)\)

\(=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)

\(1+x+y+z=1+1\Rightarrow1+x=\left(1-y\right)+\left(1-z\right)\ge2\sqrt{\left(1-y\right)\left(1-z\right)}\)

Tương tự ta cũng có: \(1+y\ge2\sqrt{\left(1-z\right)\left(1-x\right)}\)

\(1+z\ge2\sqrt{\left(1-x\right)\left(1-y\right)}\)

Vậy \(S\le\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{8\left(1-x\right)\left(1-y\right)\left(1-z\right)}=\frac{1}{8}\)

Ta sẽ chứng minh: \(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\) với a;b dương

Thật vậy, BĐT tương đương:

\(3a^3\ge\left(2a-b\right)\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)

Áp dụng: \(\Rightarrow S\ge\frac{2x-y}{3}+\frac{2y-z}{3}+\frac{2z-x}{3}=\frac{x+y+z}{3}=3\)

\(S_{min}=3\) khi \(x=y=z=3\)

Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)

\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)

Ta có \(x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2=4\Rightarrow+xy+yz+zx=-7\)

vì \(x+y+z=2\Rightarrow z-1=1-x-y\Rightarrow\frac{1}{xy+z-1}=\frac{1}{xy+1-x-y}=\frac{1}{\left(x-1\right)\left(y-1\right)}. \)

Suy ra \(S=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}. \)

               \(\frac{z-1+x-1+y-1}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}=-\frac{1}{7}\)