Ta có:
\(VT=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{xy+yz+zx}{xy}+\frac{xy+yz+zx}{yz}+\frac{xy+yz+zx}{zx}\)
\(VT=3+\frac{z\left(x+y\right)}{xy}+\frac{x\left(y+z\right)}{yz}+\frac{y\left(x+z\right)}{zx}\) (1)
Mặt khác:
\(\frac{z\left(x+y\right)}{xy}+\frac{x\left(y+z\right)}{yz}\ge2\sqrt{\frac{zx\left(x+y\right)\left(y+z\right)}{xy^2z}}=2\sqrt{\frac{\left(x+y\right)\left(y+z\right)}{y^2}}=\frac{2\sqrt{y^2+xy+yz+zx}}{y}=\frac{2\sqrt{y^2+1}}{y}\)
Tương tự: \(\frac{z\left(x+y\right)}{xy}+\frac{y\left(x+z\right)}{zx}\ge\frac{2\sqrt{x^2+1}}{x}\) ; \(\frac{x\left(y+z\right)}{yz}+\frac{y\left(x+z\right)}{zx}\ge\frac{2\sqrt{z^2+1}}{z}\)
Cộng vế với vế:
\(\frac{z\left(x+y\right)}{xy}+\frac{x\left(y+z\right)}{yz}+\frac{y\left(x+z\right)}{xz}\ge\frac{\sqrt{x^2+1}}{x}+\frac{\sqrt{y^2+1}}{y}+\frac{\sqrt{z^2+1}}{z}\) (2)
Từ (1) và (2) suy ra đpcm
Dấu "=" xảy ra khi \(x=y=z=...\)
