Ý 1: Giả sử: `x^2+y^2+z^2-xy+yz+zx=\frac{(x-y)^2+(y+z)^2+(z+x)^2}{2}`
`⇒2.(x^2+y^2+z^2-xy+yz+zx)=(x-y)^2+(y+z)^2+(z+x)^2`
`⇔2x^2+2y^2+2z^2-2xy+2yz+2zx=(x-y)^2+(y+z)^2+(z+x)^2`
`⇔(x^2-2xy+y^2)+(y^2+2yz+z^2)+(x^2+2zx+z^2)=(x-y)^2+(y+z)^2+(z+x)^2`
`⇔(x-y)^2+(y+z)^2+(z+x)^2=(x-y)^2+(y+z)^2+(z+x)^2` (luôn đúng`⇒`giả sử đúng)
Ý 2:
`x^2+y^2+z^2-xy-yz+zx=0`
`⇔2.(x^2+y^2+z^2-xy+yz+zx)=0`
`⇔2x^2+2y^2+2z^2-2xy+2yz+2zx=0`
`⇔(x^2-2xy+y^2)+(y^2+2yz+z^2)+(x^2+2zx+z^2)=0`
`⇔(x-y)^2+(y+z)^2+(z+x)^2=0`
Có: `(x-y)^2\ge0, (y+z)^2\ge0, (z+x)^2\ge0⇒(x-y)^2+(y+z)^2+(z+x)^2\ge0`
Dấu bằng xảy ra khi `x-y=0,y+z=0,z+x=0⇒x=y=-z.`
Cách khác ý 2:
Có: `x^2+y^2+z^2-xy+yz+zx=\frac{(x-y)^2+(y+z)^2+(z+x)^2}{2}` $(cmt)$
`⇒2.(x^2+y^2+z^2-xy+yz+zx)=(x-y)^2+(y+z)^2+(z+x)^2`
Mà `x^2+y^2+z^2-xy+yz+zx=0` `⇒ 0= (x-y)^2+(y+z)^2+(z+x)^2`
Có: `(x-y)^2\ge0, (y+z)^2\ge0, (z+x)^2\ge0⇒(x-y)^2+(y+z)^2+(z+x)^2\ge0`
Dấu bằng xảy ra khi `x-y=0,y+z=0,z+x=0⇒x=y=-z.`
