Đáp án:
\[14M + 2m = 12\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0,\,\,\,\forall x;y \Rightarrow {x^2} + {y^2} \ge 2xy\\
{\left( {y - z} \right)^2} \ge 0,\,\,\,\forall y;z \Rightarrow {y^2} + {z^2} \ge 2yz\\
{\left( {z - x} \right)^2} \ge 0,\,\,\,\forall z;x \Rightarrow {z^2} + {x^2} \ge 2zx\\
\Rightarrow \left( {{x^2} + {y^2}} \right) + \left( {{y^2} + {z^2}} \right) + \left( {{z^2} + {x^2}} \right) \ge 2xy + 2yz + 2zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) \ge 2\left( {xy + yz + zx} \right)\\
\Leftrightarrow xy + yz + zx \le {x^2} + {y^2} + {z^2} = 1\\
\Rightarrow M = \max \left( {xy + yz + zx} \right) = 1\\
{\left( {x + y} \right)^2} \ge 0,\,\,\,\forall x;y \Rightarrow {x^2} + {y^2} \ge - 2xy\\
{\left( {y + z} \right)^2} \ge 0,\,\,\,\forall y;z \Rightarrow {y^2} + {z^2} \ge - 2yz\\
{\left( {z + x} \right)^2} \ge 0,\,\,\,\forall z;x \Rightarrow {z^2} + {x^2} \ge - 2zx\\
\Rightarrow \left( {{x^2} + {y^2}} \right) + \left( {{y^2} + {z^2}} \right) + \left( {{z^2} + {x^2}} \right) \ge - 2xy - 2yz - 2zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) \ge - 2\left( {xy + yz + zx} \right)\\
\Leftrightarrow - \left( {xy + yz + zx} \right) \le {x^2} + {y^2} + {z^2} = 1\\
\Leftrightarrow xy + yz + zx \ge - 1\\
\Rightarrow m = \min \left( {xy + yz + zx} \right) = - 1\\
\Rightarrow 14M + 2m = 14.1 + 2.\left( { - 1} \right) = 12
\end{array}\)
