Bổ đề: \(xyz=\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\le\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Do đó:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Đặt vế trái là P
\(P=\frac{x}{x\left(x+y+z\right)+yz}+\frac{y}{y\left(x+y+z\right)+zx}+\frac{z}{z\left(x+y+z\right)+xy}=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(P=\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(P\le\frac{2\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)}=\frac{9}{4\left(x+y+z\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
