bài 2
ta có x^3+y^3+z^3=3xyz
=>x^3+y^3+z^3-3xyz=0
=>(x+y)^3-3xy(x+y)+z^3-3xyz=0
=>(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y++z)=0
=>(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=0
ta có x^2+y^2+z^2-xy-xz-yz=0
=>2(x^2+y^2+z^2-xy-xz-yz)=0
2x^2+2y^+2z^2-2xy-2xz-2yz=0
=>(x^2-2x+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)=0
(x-y)^2+(x-z)^2+(y-z)^2=0
ta có (x-y)^2>=0;(y-z)^2>=0=0
=>(x-y)^2+(x-z)^2+(y-z)^2>=0
dấu bằng xảy ra khi
x-y=0 x=y
x-z=0 x=z x=y=x(đpcm)
y-z=0 y=z