Question

x^2+y^2+z^2>=xz+yx+zy

Answer 1

(X+y)^2>=0
(X+z)^2>=0
(y+z)^2>=0
ridicam la patrat si adunam
X^2+y^2+2xy+X^2+z^2+2xz+y^2+z^2+2yz>=0

2X^2+2y^2+2z^2>=2xy+2xz+2yz /:2

X^2+y^2+z^2>=xy+xz+yz

Answer 2

x²+y²+z²≥xz+xy+yz I·2 ⇔2x²+2y²+2z²≥2xy+2xz+2yz ⇔
(x²-2xy+y²)+(x²-2xz+z²)+(y²-2yz+z²)≥0 ⇔ (x-y)²+(x-z)²+(y-z)²≥0 evident adevarata ca suma de patrate, datorita echvalentei,relatia de la care s-a plecat este adevarata.