Cho (a^2 + b^2 + c^2 = ab + bc + ca) Chứng minh rằng: (a = b = c)

Ta có \({a^2} + {b^2} + {c^2} = ab + bc + ca\)

Suy ra \(2\left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + bc + ca} \right) = 0\)

\({a^2} - 2ab + {b^2} + {b^2} - 2bc + {c^2} + {c^2} - 2ca + {a^2} = 0\)

\({\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 & (*)\)

Vì \[{\left( {a - b} \right)^2} \ge 0;\,\,\,{\left( {b - c} \right)^2} \ge 0;\,\,\,{\left( {c - a} \right)^2} \ge 0\,\,\forall a,\,\,b,\,\,c \in \mathbb{R}\] nên từ \((*)\) suy ra

\(a - b = b - c = c - a = 0\) hay \(a = b = c\).