\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=3\)
Dấu "=" \(\Leftrightarrow\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}\Leftrightarrow x=y=z=2\)
\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)=3\)
Dấu "=" xảy ra khi \(x=y=z=2\)
Cách khác
Ta có: \(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2\left(y+z\right)}{4\left(y+z\right)}}=x\)
\(\frac{y^2}{z+x}+\frac{z+x}{4}\ge2\sqrt{\frac{y^2\left(z+x\right)}{4\left(z+x\right)}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{z^2\left(x+y\right)}{4\left(x+y\right)}}=z\)
Cộng lại ta được \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge x+y+z-\frac{x+y+z}{2}=6-\frac{6}{2}=3\)
Dấu = xảy ra khi \(x=y=z=2\)
