\(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)\(\Leftrightarrow x=y=z\)
Ta có : \(xy+yz+xz\le x^2+y^2+z^2\)
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\) ( nhân 2 vế cho 2 )
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) vì \(\left(x-y\right)^2\ge0,\left(y-z\right)^2\ge0,\left(x-z\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\Leftrightarrow x=y=z=0}\)
