Áp dụng bđt cosi ta có
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
Suy ra \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng ta có \(\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge9\Leftrightarrow2\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge9\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{9}{2}\Leftrightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}\ge\frac{9}{2}\Leftrightarrow\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}+3\ge\frac{9}{2}\Leftrightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{3}{2}\)
Vậy GTNN của \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\) là \(\frac{3}{2}\)
Áp dụng BĐT Cauchy - schwarz dưới dạng en-gel ta có :
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{x^2}{xy+zx}+\frac{y^2}{yz+xy}+\frac{z^2}{zx+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{\frac{2\left(x+y+z\right)^2}{3}}=\frac{3}{2}\)
Vậy GTNN của biểu thức là \(\frac{3}{2}\) . Dấu \("="\) xảy ra khi \(x=y=z\)