Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0,\,\,\,\,\forall a,b,c\\
\Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) \ge 0\\
\Leftrightarrow 2.\left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + bc + ca} \right) \ge 0\\
\Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) \ge 2.\left( {ab + bc + ca} \right)\\
\Leftrightarrow 3\left( {{a^2} + {b^2} + {c^2}} \right) \ge \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {ab + bc + ca} \right)\\
\Leftrightarrow 3\left( {{a^2} + {b^2} + {c^2}} \right) \ge {\left( {a + b + c} \right)^2}\\
\Rightarrow \frac{{{{\left( {a + b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2}}} \le 3\\
\Rightarrow {y_{\max }} = 3 \Leftrightarrow a = b = c
\end{array}\)
