\(A=\frac{x^2}{xy+2xz}+\frac{y^2}{zy+2xy}+\frac{z^2}{xz+2yz}\)
\(A\ge\frac{\left(x+y+z\right)^2}{3\left(xy+yz+zx\right)}\left(cauchy-schwarz\right)\)
Sử dụng đánh giá quen thuộc:\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge1\)
"="<=>x=y=z