Ta có: a2 + b2 + c2 = ab + bc + ca
2(a2 + b2 + c2) = 2(ab + bc + ca)
2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
(a2 − 2ab + b2) + (b2 − 2bc + c2) + (c2 − 2ca + a2) = 0
(a − b)2 + (b − c)2 + (c − a)2 = 0
Mà (a − b)2 ≥ 0; (b − c)2 ≥ 0; (c − a)2 ≥ 0 nên suy ra
Ta có : a2+b2+c2=ab+bc+ca
⇔2(a2+b2+c2)=2(ab+bc+ca)
⇔2a2+2b2+2c2- 2ab-2bc-2ca=0
⇔(a2- 2ab+b2)+(b2-2bc+c2)+(c2-2ca+a2)=0
⇔(a-b)2+(b-c)2+(c-a)2=0
a-b=0 a=b
b-c=0 ⇔ b=c
c-a=0 c=a
⇔a=b=c (đpcm)