求证:
$$
\frac{n}{\sum\limits_{i = 1} ^ {n} \dfrac{1}{x_i}} \le \sqrt [n] {\prod_{i = 1} ^ {n} x_i } \le \frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \le \sqrt{\frac{\sum\limits_{i = 1} ^ {n} x_i ^ 2}{n}}
$$
分析:
引理:琴生不等式
若 $f(x)$ 是区间 $[a,b]$ 的凹函数,则对任意 $x_1,x_2,\cdots,x_n \in [a,b]$ 有不等式
$$
f\left(\frac{\sum\limits_{i = 1} ^ {n} x_i}{n}\right) \le \frac{\sum\limits_{i = 1} ^ {n} f(x_i)}{n}
$$
若 $f(x)$ 是区间 $[a,b]$ 的凸函数,则对任意 $x_1,x_2,\cdots,x_n \in [a,b]$ 有不等式
$$
\frac{\sum\limits_{i = 1} ^ {n} f(x_i)}{n}\le f\left(\frac{\sum\limits_{i = 1} ^ {n} x_i}{n}\right)
$$
设 $f(x) = \ln x$,易得 $\ln x$ 为凸函数,所以有
$$ \frac{\sum\limits_{i = 1} ^ {n} \ln x_i}{n} \le \ln \frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \\\\ \Leftrightarrow \frac{\ln \prod\limits_{i = 1} ^ {n} x_i}{n} \le \ln \frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \\\\ \Leftrightarrow \ln \prod\limits_{i = 1} ^ {n} x_i \le \ln \left (\frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \right) ^ n \\\\ \Leftrightarrow \prod\limits_{i = 1} ^ {n} x_i \le \left (\frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \right) ^ n \\\\ \Leftrightarrow \sqrt[n]{\prod\limits_{i = 1} ^ {n} x_i} \le \frac{\sum\limits_{i = 1} ^ {n} x_i}{n} $$
第二个不等式得证。
若对于第二个不等式做变换 $x_i \rightarrow \dfrac{1}{x_i}$,有
$$ \sqrt[n]{\prod\limits_{i = 1} ^ {n} \frac{1}{x_i}} \le \frac{\sum\limits_{i = 1} ^ {n} \dfrac{1}{x_i}}{n} \\\\ \Leftrightarrow \frac{1}{\sqrt[n]{\prod\limits_{i = 1} ^ {n} x_i}} \le \frac{\sum\limits_{i = 1} ^ {n} \dfrac{1}{x_i}}{n} \\\\ \Leftrightarrow \frac{n}{\sum\limits_{i = 1} ^ {n} \dfrac{1}{x_i}} \le \sqrt[n]{\prod\limits_{i = 1} ^ {n} x_i} $$
第一个不等式得证。
再设 $f(x) = x ^ 2$,易得 $f(x)$ 在 $x \in [0, \infty]$ 为凹函数,所以有
$$
\left(\frac{\sum\limits_{i = 1} ^ {n} x_i}{n}\right) ^ 2 \le \frac{\sum\limits_{i = 1} ^ {n} x_i ^ 2}{n} \\\\
\Leftrightarrow \frac{\sum\limits_{i = 1} ^ {n} x_i}{n} \le \sqrt{\frac{\sum\limits_{i = 1} ^ {n} x_i ^ 2}{n}}
$$
第三个不等式得证。