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