Theo mình thì bài này làm như sau
Ta có: \(a^2+b^2+c^2=ab+bc+ac\)
\(\begin{array}{l} \Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) = 2\left( {ab + bc + ca} \right)\\ \Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \end{array}\)
Vì \({\left( {a - b} \right)^2} \ge 0;\,\,{\left( {b - c} \right)^2} \ge 0;\,\,{\left( {c - a} \right)^2} \ge 0\) nên \({\left( {a - b} \right)^2} + \,\,{\left( {b - c} \right)^2} + \,\,{\left( {c - a} \right)^2} \ge 0\) khi
\(\left\{ \begin{array}{l} a - b = 0\\ b - c = 0\\ c - a = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = b\\ b = c\\ c = a \end{array} \right. \Leftrightarrow a = b = c\ (đpcm)\)
