均值不等式证明(拉格朗日乘数法)

Part 1

求证:\(\frac{n}{\sum_{i=1}^{n}\frac{1}{y_i}} \leq ({\prod_{i=1}^{n}y_i})^{\frac{1}{n}}\)

\(y_i\) 为正实数

\(n \geq 3\)

证明:

\(x_i\) = \(y_i^{\frac{1}{n}}\)

\(x_i\) 为正实数

原命题等价于:

\[{\prod_{i=1}^{n}x_i} - \frac{n}{\sum_{i=1}^{n}\frac{1}{x_i^{n}}} \geq 0 \]

定义函数:

\[f(x_1,x_2...x_n) = {\prod_{i=1}^{n}x_i} - \frac{n}{\sum_{i=1}^{n}\frac{1}{x_i^{n}}} \]

有:

\[f_{x_i}(x_1,x_2...x_n)' = \prod_{j=1,j \not = i}^{n}x_j - n^2 (\sum_{j=1}^{n} x_j^{-n})^{-2} x_i^{-n-1} \]

又因为取到极值时,所有偏导等于 \(0\)

所以:

\[\prod_{j=1,j \not = i}^{n}x_j = n^2 (\sum_{j=1}^{n} x_j^{-n})^{-2} x_i^{-n-1} \]

简单变形得:

\[x_i^{n} = \frac{n^2 (\sum_{j=1}^{n} x_j^{-n})^{-2}}{\prod_{j=1}^{n}x_j} \]

所以:

\[x_1^{n} = x_2^{n} =...= x_n^{n} \]

又因为 \(x_i\) 为正实数

所以:

\[x_1 = x_2 =...= x_n \]

我们设这个极值在 \((x_0,x_0...x_0)\) 处取到

又因为:

\[f_{x_i}(x_0,x_0...x_0)'' = x_0^{n-2} \]

且:

\[x_0 > 0 \]

所以:

在此处 \(f\) 取到最小值

即:

\(f \geq 0\)

故命题得证

Part 2

求证:

\[({\prod_{i=1}^{n}y_i})^{\frac{1}{n}} \leq \sum_{i=1}^{n} \frac{y_i}{n} \]

\(y_i\) 为正实数(其实这里也可以是非负实数,但是这个问题留给读者自证)

\(n \geq 3\)

证明:

\(x_i\) = \(y_i^{\frac{1}{n}}\)

\(x_i\) 为正实数

原命题等价于:

\[\sum_{i=1}^{n} x_i ^n \geq n \prod_{i=1}^{n} x_i \]

依然定义:

\[f(x1,x2...x_n) = \sum_{j=1}^{n} x_j ^n - n \prod_{j=1}^{n} x_j \]

有:

\[f_{x_i}'(x_1,x_2...,x_n) = n x_j^{n-1} - n \prod_{j=1,j \not = i}^{n} x_j \]

依然在极值处取到 \(0\)

所以:

\[n x_j^{n-1} = n \prod_{j=1,j \not = i}^{n} x_j \]

所以:

\[x_j^{n} = \prod_{j=1}^{n} x_j \]

所以:

\[x_1^{n} = x_2^{n} =...= x_n^{n} \]

又因为 \(x_i\) 为正实数

所以:

\[x_1 = x_2 =...= x_n \]

又因为:

\[f_{x_i}(x_0,x_0...x_0)'' = (n^2 - n) x_0^{n-2} \]

且:

\(x_0 > 0\)

所以:

在此处 \(f\) 取到最小值

即:

\(f \geq 0\)

故命题得证

Part 3

求证:

\[\sqrt{\frac{\sum_{i=1}^{n}x_i^2}{n}} \geq \frac{\sum_{i=1}^{n}x_i}{n} \]

\(x_i\) 为实数

\(n \geq 3\)

证明:

简单推导可知原命题等价于:

\[n \sum_{i=1}^{n} x_i^2 \geq (\sum_{i=1}^{n}x_i)^2 \]

定义函数:

\[f(x_1,x_2...x_n) = n \sum_{i=1}^{n} x_i^2 -(\sum_{i=1}^{n}x_i)^2 \]

有:

\[f_{x_i}'(x_1,x_2...x_n) = 2 n x_i - 2 (\sum_{j=1}^{n}x_j) \]

让其取到 \(0\)

有:

\[x_i = \frac{\sum_{j=1}^{n}x_j}{n} \]

所以:

\[x_1 = x_2 =...= x_n \]

又因为:

\[f_{x_i}''(x_1,x_2...x_n) = 2 n - 2 \]

且:

\(2 n - 2 > 0\)

所以:

\(f\) 只会取到最小值

有:

\(f \geq 0\)

命题得证

至此,命题全部证完

posted @ 2024-02-27 18:13  ChiFAN鸭  阅读(318)  评论(0编辑  收藏  举报