\(x+y+z=6\)
\(\Rightarrow\left(x+y+z\right)^2=36\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=36\)
\(\Rightarrow2xy+2yz+2zx=24\)
\(\Rightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y-z\right)^2\ge0\forall y,z\\\left(z-x\right)^2\ge0\forall z,x\end{matrix}\right.\)
Nên \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-y\right)^2=\left(y-z\right)^2=\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow x=y=z=2\)