$\frac{x^2}{y^2}+1\ge2|\frac xy|\geq 2\frac{x}{y}tuongtuconglaiVT\ge 2(\frac xy+\frac yz+\frac zx)-3\ge VP$ (AM-GM 3 so)
x2y2+1≥2|xy|tuong tu cong lai$VT\ge 2(
|\frac xy
|+
|\frac yz
|+
|\frac zx
|)-3
\ge \frac xy + \frac yz + \frac zx+(|\frac xy|+|\frac yz|+|\frac zx|-3)\ge VP$ (AM-GM 3 so)