\(a^2+b^2+c^2=ab+bc+ca\)
\(=>a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)
\(< =>a-b=0,b-c=0,c-a=0\)
\(=>a=b,b=c,c=a\)
Vậy \(a=b=c\)