高中数学选择性必修 第二册（381题）

---

知识点：第四章 数列

参考答案：猜想 \(\left( {{x_1} + {x_2} + \cdots + {x_n}} \right)\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_n}}}} \right) \geqslant {n^2}\,\left( {n \geqslant 2,\,n \in {N^*}} \right)\) ，
证明如下：①当 \(n = 2\) 时，由已知得猜想成立；
②假设当 \(n = k\) 时，猜想成立，即 \(\left( {{x_1} + {x_2} + \cdots + {x_k}} \right)\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_k}}}} \right) \geqslant {k^2}\) ， 则当 \(n = k + 1\) 时， \(\left( {{x_1} + {x_2} + \cdots + {x_k} + {x_{k + 1}}} \right)\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_k}}} + \frac{1}{{{x_{k + 1}}}}} \right)\) \( = \left( {{x_1} + {x_2} + \cdots + {x_k}} \right)\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_k}}}} \right) + \left( {{x_1} + {x_2} + \cdots + {x_k}} \right)\frac{1}{{{x_{k + 1}}}} + {x_{k + 1}}\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_k}}}} \right) + 1\) \( \geqslant {k^2} + \left( {{x_1} + {x_2} + \cdots + {x_k}} \right)\frac{1}{{{x_{k + 1}}}} + {x_{k + 1}}\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} + \cdots + \frac{1}{{{x_k}}}} \right) + 1\) \( = {k^2} + \left( {\frac{{{x_1}}}{{{x_{k + 1}}}} + \frac{{{x_{k + 1}}}}{{{x_1}}}} \right) + \left( {\frac{{{x_2}}}{{{x_{k + 1}}}} + \frac{{{x_{k + 1}}}}{{{x_2}}}} \right) + \cdots + \left( {\frac{{{x_2}}}{{{x_{k + 1}}}} + \frac{{{x_{k + 1}}}}{{{x_2}}}} \right) + 1\) ，\( \geqslant {k^2} + 2 + 2 + \cdots + 2 + 1 = {k^2} + 2k + 1 = {\left( {k + 1} \right)^2}\) 所以当 \(n = k + 1\) 时原式成立.
综合①②可知，猜想成立.