Ta có:$S=\frac{x^{2}.(1-y)}{y}+\frac{y^{2}(1-z)}{z}+\frac{z^{2}.(1-x)}{x}-(x^{2}+y^{2}+z^{2}).$Vì $00,1-y>0\rightarrow \frac{x^{2}.(1-y)}{y}>0,y.(1-y)>0$Áp dụng bđt Cosi ta có:$\frac{x^{2}.(1-y)}{y}+y(1-y)\geq 2x(1-y)$Tương tự rồi cộng vế vs vế ta đc $S\geq 2x-2xy+2y-2yz+2z-2xz-x-y-z=(x+y+z)-2(xy+yz+zx)=(x+y+z)-2$Mặt khác :$x^{2}+y^{2}+z^{2}\geq xy+yz+zx$$\rightarrow (x+y+z)^{2}\geq 3.(xy+yz+zx)\rightarrow x+y+z\geq \sqrt{3.(xy+yz+zx)}=\sqrt{3}$-->$S\geq \sqrt{3}-2$.Dấu "=" xảy ra $\leftrightarrow \begin{cases}x= y=z\\ xy+yz+zx=1 \end{cases}\leftrightarrow x=y=z=\frac{1}{\sqrt{3}}$

Ta có:$S=\frac{x^{2}.(1-y)}{y}+\frac{y^{2}(1-z)}{z}+\frac{z^{2}.(1-x)}{x}-(x^{2}+y^{2}+z^{2}).$Vì$0<x,y,z&lt;1\rightarrow y>0,1-y>0$Áp dụng bđt Cosi ta có:$\frac{x^{2}.(1-y)}{y}+y(1-y)\geq 2x(1-y)$Tương tự rồi cộng vế vs vế ta đc $S\geq 2x-2xy+2y-2yz+2z-2xz-x-y-z=(x+y+z)-2(xy+yz+zx)=(x+y+z)-2$Mặt khác :$x^{2}+y^{2}+z^{2}\geq xy+yz+zx$$\rightarrow (x+y+z)^{2}\geq 3.(xy+yz+zx)\rightarrow x+y+z\geq \sqrt{3.(xy+yz+zx)}=\sqrt{3}$-->$S\geq \sqrt{3}-2$.Dấu "=" xảy ra $\leftrightarrow \begin{cases}x= y=z\\ xy+yz+zx=1 \end{cases}\leftrightarrow x=y=z=\frac{1}{\sqrt{3}}$

Ta có:$S=\frac{x^{2}.(1-y)}{y}+\frac{y^{2}(1-z)}{z}+\frac{z^{2}.(1-x)}{x}-(x^{2}+y^{2}+z^{2}).$Vì $0<x,y,z<1\rightarrow y>0,1-y>0\rightarrow \frac{x^{2}.(1-y)}{y}>0,y.(1-y)>0$Áp dụng bđt Cosi ta có:$\frac{x^{2}.(1-y)}{y}+y(1-y)\geq 2x(1-y)$Tương tự rồi cộng vế vs vế ta đc $S\geq 2x-2xy-+2y-2yz+2z-2xz-x-y-z=(x+y+z)-2(xy+yz+zx)=(x+y+z)-2$Mặt khác :$x^{2}+y^{2}+z^{2}\geq xy+yz+zx$$\rightarrow (x+y+z)^{2}\geq 3.(xy+yz+zx)\rightarrow x+y+z\geq \sqrt{3.(xy+yz+zx)}=\sqrt{3}$-->$S\geq \sqrt{3}-2$.Dấu "=" xảy ra $\leftrightarrow \begin{cases}x= y=z\\ xy+yz+zx=1 \end{cases}\leftrightarrow x=y=z=\frac{1}{\sqrt{3}}$

Ta có:$S=\frac{x^{2}.(1-y)}{y}+\frac{y^{2}(1-z)}{z}+\frac{z^{2}.(1-x)}{x}-(x^{2}+y^{2}+z^{2}).$Vì $00,1-y>0\rightarrow \frac{x^{2}.(1-y)}{y}>0,y.(1-y)>0$Áp dụng bđt Cosi ta có:$\frac{x^{2}.(1-y)}{y}+y(1-y)\geq 2x(1-y)$Tương tự rồi cộng vế vs vế ta đc $S\geq 2x-2xy+2y-2yz+2z-2xz-x-y-z=(x+y+z)-2(xy+yz+zx)=(x+y+z)-2$Mặt khác :$x^{2}+y^{2}+z^{2}\geq xy+yz+zx$$\rightarrow (x+y+z)^{2}\geq 3.(xy+yz+zx)\rightarrow x+y+z\geq \sqrt{3.(xy+yz+zx)}=\sqrt{3}$-->$S\geq \sqrt{3}-2$.Dấu "=" xảy ra $\leftrightarrow \begin{cases}x= y=z\\ xy+yz+zx=1 \end{cases}\leftrightarrow x=y=z=\frac{1}{\sqrt{3}}$