x2+y2+z2+3> hoac = 2(x+y+z)
\(x^2+y^2+z^2+3-2\left(x+y+z\right)\ge0\)
\(\Rightarrow x^2+y^2+z^2+3-2x-2y-2z\ge0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(z^2-2z+1\right)\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)(Đpcm)
Dấu = khi (x-1)2=(y-1)2=(z-1)2=0 =>x=y=z=1