Question

Sa se arate ca daca $x^{2} + y^{2} +z^{2}=xy+yz+xz$, atunci x=y=z, x,y,z ∈R.

Answer 1

inmultim ex . cu 2  
2x² + 2y² + 2z²  = 2xy + 2yz + 2xz 
2x²  +      2y² +      2z² - 2xy - 2yz - 2xz = 0 
↓               ↓           ↓
x²+x²       y²+y²         z²+z²       formam binoame 
( x² - 2xy +y² ) + ( x² -2xz + z² ) + ( y² -2yz + z² ) = 0 
( x -y ) ² + ( x -z ) ² + ( y -z ) ² =0 suma de patrate este nula daca  : 
x -y =0        x -z =0        y -z = 0 
x =y =z

Answer 2

[tex]\frac{x^2+y^2}{2}\geq\sqrt{x^2\cdot y^2}=|xy|\geq xy,\ \forall\ x,y\in\mathbb{R}\\ \text{Egalitatea are loc daca si numai daca }x=y\\ \frac{x^2+y^2}{2}\geq xy\\ \frac{y^2+z^2}{2}\geq yz\\ \frac{x^2+z^2}{2}\geq xz\\ ................................+\\ \frac{x^2+y^2}{2}+\frac{y^2+z^2}{2}+\frac{z^2+x^2}{2}=x^2+y^2+z^2\geq xy+yz +xz\\ \text{Egalitatea are loc daca si numai daca }x=y,y=z \text{ si $x=z$ adica }\\ x=y=z.[/tex]