bài 2:$VT=\frac{x^2}{xy+2xz}+\frac{y^2}{yz+2xy}+\frac{z^2}{xz+2yz}\geq \frac{(x+y+z)^2}{xy+2xz+yz+2xy+xz+2yz}= \frac{(x+y+z)^2}{3(xy+yz+zx)}\geq \frac{(x+y+z)^2}{(x+y+z)^2}=1\Rightarrow Min=1$
bài 2:$
\color{pink}{VT=\frac{x^2}{xy+2xz}+\frac{y^2}{yz+2xy}+\frac{z^2}{xz+2yz}\geq \frac{(x+y+z)^2}{xy+2xz+yz+2xy+xz+2yz}= \frac{(x+y+z)^2}{3(xy+yz+zx)}\geq \frac{(x+y+z)^2}{(x+y+z)^2}=1\Rightarrow Min=1
}$