\(x+y+z+xy+yz+zx=\frac{9}{4}\)
Ta có:
\(4x^2+1+4y^2+1+4z^2+1\ge4x+4y+4z\)
\(4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)\)
Cộng vế với vế:
\(8\left(x^2+y^2+z^2\right)+3\ge4\left(x+y+z+xy+yz+zx\right)\)
\(\Leftrightarrow8\left(x^2+y^2+z^2\right)+3\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge\frac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)