Đáp án:
\[\frac{{100}}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0 \Rightarrow {x^2} - 2xy + {y^2} \ge 0 \Rightarrow {x^2} + {y^2} \ge 2xy\\
{\left( {y - z} \right)^2} \ge 0 \Rightarrow {y^2} - 2yz + {z^2} \ge 0 \Rightarrow {y^2} + {z^2} \ge 2yz\\
{\left( {z - x} \right)^2} \ge 0 \Rightarrow {z^2} - 2zx + {x^2} \ge 0 \Rightarrow {z^2} + {x^2} \ge 2zx\\
\Rightarrow \left( {{x^2} + {y^2}} \right) + \left( {{y^2} + {z^2}} \right) + \left( {{z^2} + {x^2}} \right) \ge 2xy + 2yz + 2zx\\
\Rightarrow {x^2} + {y^2} + {z^2} \ge xy + yz + zx\\
x + y + z = 1\\
\Leftrightarrow {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx = 1\\
\Leftrightarrow 1 \le {x^2} + {y^2} + {z^2} + 2\left( {{x^2} + {y^2} + {z^2}} \right) = 3\left( {{x^2} + {y^2} + {z^2}} \right)\\
\Rightarrow {x^2} + {y^2} + {z^2} \ge \frac{1}{3}\\
1 = x + y + z \ge 3.\sqrt[3]{{x.y.z}} \Rightarrow xyz \le \frac{1}{{27}}\\
M = {\left( {x + \frac{1}{x}} \right)^2} + {\left( {y + \frac{1}{y}} \right)^2} + {\left( {z + \frac{1}{z}} \right)^2}\\
= {x^2} + 2x.\frac{1}{x} + \frac{1}{{{x^2}}} + {y^2} + 2.y.\frac{1}{y} + \frac{1}{{{y^2}}} + {z^2} + 2.z.\frac{1}{z} + \frac{1}{{{z^2}}}\\
= \left( {{x^2} + {y^2} + {z^2}} \right) + \left( {\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}}} \right) + 6\\
\ge \frac{1}{3} + 3.\sqrt[3]{{\frac{1}{{{x^2}}}.\frac{1}{{{y^2}}}.\frac{1}{{{z^2}}}}} + 6\\
\ge \frac{1}{3} + 3.\sqrt[3]{{\frac{1}{{\frac{1}{{{{27}^2}}}}}}} + 6 = \frac{1}{3} + 27 + 6 = \frac{{100}}{3}
\end{array}\)
