Theo cosi tao có: \(x^2+1\ge2x\Rightarrow x^2\ge2x-1\left(1\right)\) ; \(y^2+1\ge2y\Rightarrow y^2\ge2y-1\left(2\right)\)\(z^2+1\ge2z\Rightarrow z^2\ge2z-1\left(3\right)\)Lại có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+xz+yz\right)\ge0\)\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+xz+yz\right)\left(4\right)\)Cộng vế theo vế của (1) (2) (3) (4) ta...
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Theo cosi tao có: \(x^2+1\ge2x\Rightarrow x^2\ge2x-1\left(1\right)\) ; \(y^2+1\ge2y\Rightarrow y^2\ge2y-1\left(2\right)\)
\(z^2+1\ge2z\Rightarrow z^2\ge2z-1\left(3\right)\)
Lại có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+xz+yz\right)\ge0\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+xz+yz\right)\left(4\right)\)
Cộng vế theo vế của (1) (2) (3) (4) ta có \(3\left(x^2+y^2+z^2\right)\ge2\left(x+y+z+xy+xz+yz\right)-3=2.6-3=9\)
\(\Rightarrow x^2+y^2+z^2\ge3\left(dpcm\right)\)
