Có:
`a^2 + b^2 + c^2 = ab + bc +ca`
`⇔2.(a^2 + b^2 + c^2) = 2.(ab + bc +ca)`
`⇔2a^2 + 2b^2 + 2c^2 = 2ab + 2bc +2ca`
`⇔2a^2 + 2b^2 + 2c^2 - (2ab + 2bc +2ca)=0`
`⇔2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca=0`
`⇔(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0`
`⇔(a-b)^2+(b-c)^2+(c-a)^2=0`
Có: `(a-b)^2 \ge 0 , (b-c)^2 \ge 0, (c-a)^2 \ge 0 `
`⇒(a-b)^2+(b-c)^2+(c-a)^2 \ge 0`
Dấu ''='' xảy ra khi:
`(a-b)^2=0, (b-c)^2=0, (c-a)^2=0`
`⇔a-b=0, b-c=0, c-a=0`
`⇔a=b=c.`
Thay `a=b=c` vào `A = (a-b)^{2015} + (b-c)^{2016} + (c-a)^{2017}`, ta có:
`A = (a-a)^{2015} + (b-b)^{2016} + (c-c)^{2017}`
`A=0.`
Vậy với `a,b,c∈RR, a^2 + b^2 + c^2 = ab + bc +ca ⇒ A=0.`
