cũng bằng 3

Ta coˊ :xy+x+1x​+yz+y+1y​+xz+z+1z​

=���+�+1+�����+��+�+����2��+���+��=xy+x+1x​+xyz+xy+xxy​+x2yz+xyz+xyxyz​

=���+�+1+����+�+1+1��+�+1(Vıˋ ���=1)=xy+x+1x​+xy+x+1xy​+xy+x+11​(Vıˋ xyz=1)

=�+��+1��+�+1=xy+x+1x+xy+1​

$=1$

Ta có : \(\frac{x^3}{z+x^2}=\frac{x^3+xz-xz}{z+x^2}=x-\frac{xz}{z+x^2}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\ge x-\frac{z+1}{4}\) (Cosi)

Tương tự \(\hept{\begin{cases}\frac{y^3}{x+y^2}\ge y-\frac{x+1}{4}\\\frac{z^3}{y+z^2}\ge z-\frac{y+1}{4}\end{cases}}\)

\(\Rightarrow\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\)

Mà \(xy+yz+xz=3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\Rightarrow x+y+z\ge3\)

\(\Rightarrow\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

bước cuối sai \(\frac{3}{2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) trong khi \(3\le x+y+z\) ?? :D

à hay t sai ấy nhỉ :)) 

Ta có \(xy+yz+xz=\frac{2^2-18}{2}=-7\)

 \(x+y+z=2\)=> \(z-1=-x-y+1\)

=> \(\frac{1}{xy+z-1}=\frac{1}{xy-x-y+1}=\frac{1}{\left(x-1\right)\left(y-1\right)}\)

Tương tự \(\frac{1}{yz+x-1}=\frac{1}{\left(y-1\right)\left(z-1\right)};\frac{1}{xz+y-1}=\frac{1}{\left(z-1\right)\left(x-1\right)}\)

=> \(S=\frac{x+y+z-3}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=-\frac{1}{xyz-\left(yz+xy+xz\right)+\left(x+y+z\right)-1}\)

                                                                 \(=\frac{-1}{-1+7+2-1}=-\frac{1}{7}\)

Vậy \(S=-\frac{1}{7}\)