Question

[💥Hoàng Thị Diệu Thùy 💦]

cmr: \(x^2+y^2+z^2\ge xy-xz+yz\forall x,y,z\)

Answer 1

[Phạm Thị Thùy Linh]

\(x^2+y^2+z^2\ge xy-xz+yz\)

\(\Rightarrow2x^2+2y^2+2z^2\ge2xy-2xz+2yz\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy+2xz-2yz\ge0\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(x^2+2xz+z^2\right)+\left(z^2-2yz+y^2\right)\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(x+z\right)^2+\left(z-y\right)^2\ge0\)( luôn đúng )

\(\Rightarrow x^2+y^2+z^2\ge xy-xz+yz\)( đúng với mọi x,y,z )

Dấu bằng sảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+z\right)^2=0\\\left(z-y\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\x+z=0\\z-y=0\end{cases}\Rightarrow\hept{\begin{cases}y=x\\x+z=0\\y=z\end{cases}}}}\)

\(\Rightarrow\hept{\begin{cases}x+z=0\\x=z\end{cases}\Rightarrow x=y=z=0}\)