$A=\sum_{}^{}\frac{(x-1)^2}{z}\geq \frac{(x+y+z-3)^2}{x+y+z}=\frac{1}{2}$.(Vì $x+y+z=2$)Dấu $=$ xảy ra khi $\frac{(x-1)^2}{z}=\frac{(y-1)^2}{x}=\frac{(z-1)^2}{y}\Leftrightarrow x=y=z=\frac{2}{3}$
$A=\sum_{}^{}\frac{(x-1)^2}{z}\geq \frac{(x+y+z-3)^2}{x+y+z}=\frac{1}{2}$.(Vì $x+y+z=2$)Dấu $=$ xảy ra khi $\frac{(x-1)^2}{z}=\frac{(y-1)^2}{x}=\frac{(x-1)^2}{y}\Leftrightarrow x=y=z=\frac{2}{3}$
$A=\sum_{}^{}\frac{(x-1)^2}{z}\geq \frac{(x+y+z-3)^2}{x+y+z}=\frac{1}{2}$.(Vì $x+y+z=2$)Dấu $=$ xảy ra khi $\frac{(x-1)^2}{z}=\frac{(y-1)^2}{x}=\frac{(
z-1)^2}{y}\Leftrightarrow x=y=z=\frac{2}{3}$