Ta có: a² + b² + c² = ab + bc + ca

Û 2(a² + b² + c²) = 2(ab + bc + ca)

Û 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

Û (a² − 2ab + b²) + (b² − 2bc + c²) + (c² − 2ca + a²) = 0

Û (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

$\left\{\begin{array}{l}{\left(a-b\right)}^2=0\\ {\left(b-c\right)}^2=0\\ {\left(c-a\right)}^2=0\end{array}\right.\iff \left\{\begin{array}{l}a=b\\ b=c\\ c=a\end{array}\right.\iff a=b=c$ (đpcm).