\(x+y+z=1\)
\(< =>\left(x+y+z\right)^2=1^2\)
\(x^2+y^2+z^2+xy+yz+xz=1\)
\(x^2+y^2+z^2>=xy+yz+xz\)(Bất đẳng thức cô-si)
\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=1\)
vì \(x^2+y^2+z^2>=xy+yz+xz\)
\(< =>x^2+y^2+z^2+2\left(x^2+y^2+z^2\right)>=1\)
\(3\left(x^2+y^2+z^2\right)>=1\)
\(< =>x^2+y^2+z^2>=\frac{1}{3}\)
DPCM