Ta có: [tex]\frac{x^2}{y-1}+\frac{y^2}{z-1}+\frac{z^2}{x-1}\geq \frac{(x+y+z)^2}{x+y+z-3}[/tex] (BĐT Cauchy-Schwarz) Đặt $x+y+z=a \ge 3$ Khi đó, $\frac{a^2}{a-3}=\frac{a^2-12a+36+12a-36}{a-3}$ $=\frac{(a-6)^2+12(a-3)}{a-3}=\frac{(a-6)^2}{a-3}+12\ge 12$ Dấu "=" xảy ra khi $x=y=z=2$
