Question

Aratati ca x²+y²+z²≥xy+xz+yz , oricare x y z apartin R

Answer 1

x²+y²+z²≥xy+xz+yz
x²+y²+z²-2xy-2xz-2yz+xy+xz+yz≥0
(x-y-z)²+xy+xz+yz≥0
(x-y-z)²≥0
xy+xz+yz≥0 acolada=> propoziția e adevarata

Answer 2

[tex]\it x^2+y^2+z^2\geq xy+yz+zx |_{\cdot2} \Leftrightarrow 2x^2+2y^2+2z^2\geq 2xy+2yz+2zx \\\;\\ \Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2 -2xy-2yz-2zx \geq 0 \Leftrightarrow \\\;\\ \Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2+2zx+x^2) \geq 0 \Leftrightarrow \\\;\\ \Leftrightarrow (x-y)^2+(y-x)^2+(z-x)^2\geq0\ \ (A)[/tex]