Sửa đề:
\(\dfrac{x^2y}{x-1}+\dfrac{y^2z}{y-1}+\dfrac{z^2x}{z-1}=\dfrac{x^2y^2}{xy-y}+\dfrac{y^2z^2}{yz-z}+\dfrac{z^2x^2}{zx-x}\)
\(\ge\dfrac{\left(xy+yz+zx\right)^2}{xy+yz+zx-6}\)
Đặt \(t=xy+yz+zx>x+y+z=6\) thì ta có
\(\dfrac{t^2}{t-6}=24+\dfrac{t^2-24t+144}{t-6}=24+\dfrac{\left(t-12\right)^2}{t-6}\ge24\)
Vậy GTNN là 24 đạt dược khi \(x=y=z=2\)