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Jensen不等式导出著名不等式

分析笔记 huijiaorz 591℃ 0评论

常用的著名不等式,从Jensen不等式出发导出其他一些知名不等式

加权AG不等式

对$a_i>0,\alpha_i>0$有
$$\frac{\alpha_1a_1+\alpha_2a_2+\cdots+\alpha_na_n}{\alpha_1+\alpha_2+\cdots+\alpha_n}\geq(a_1^{\alpha_1}a_2^{\alpha_2}\cdots{a_n^{\alpha_n}})^{\frac{1}{\alpha_1+\alpha_2+\cdots+\alpha_n}}$$

证明:记$$\lambda_i=\frac{\alpha_i}{\alpha_1+\alpha_2+\cdots+\alpha_n}$$

因为对数函数为凸函数,使用加权琴生不等式,可得
$$\begin{aligned}\ln(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n)&\geq\lambda_1\ln{x_1}+\lambda_2\ln{x_2}+\cdots+\lambda_n\ln{x_n}\\&=\ln{x_1^{\lambda_1}x_2^{\lambda_2}\cdots{x}_n^{\lambda_n}}\\\therefore\quad\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n&\geq{x_1^{\lambda_1}x_2^{\lambda_2}\cdots{x}_n^{\lambda_n}}\\\therefore\quad\frac{\alpha_1a_1+\alpha_2a_2+\cdots+\alpha_na_n}{\alpha_1+\alpha_2+\cdots+\alpha_n}&\geq(a_1^{\alpha_1}a_2^{\alpha_2}\cdots{a_n^{\alpha_n}})^{\frac{1}{\alpha_1+\alpha_2+\cdots+\alpha_n}}\end{aligned}$$

Young不等式

若$$x>0,y>0\\p>1,q>1,\frac{1}{p}+\frac{1}{q}=1$$

则$$xy\leq\frac{x^p}{p}+\frac{y^q}{q}$$

证明:利用上述“加权AG不等式”有
$$x_1^{\frac{\alpha_1}{\alpha_1+\alpha_2}}\cdot{x_2}^{\frac{\alpha_1}{\alpha_1+\alpha_2}}\leq\frac{\alpha_1{x_1}+\alpha_2{x_2}}{\alpha_1+\alpha_2}$$


$$\frac{1}{p}=\frac{\alpha_1}{\alpha_1+\alpha_2}\\\frac{1}{q}=\frac{\alpha_2}{\alpha_1+\alpha_2}$$

带入整理可得Young不等式。

AG不等式

假设上述“加权AG不等式”中$\alpha_i=1$则得AG不等式$$\frac{a_1+a_2+\cdots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots{a_n}}$$

哈代不等式

若对下列正数$$x_1,x_2,\cdots,x_n\\y_1,y_2,\cdots,y_n\\\alpha,\beta,\frac{1}{\alpha}+\frac{1}{\beta}=1$$

成立不等式:$$\sum_{k=1}^nx_ky_k\leq(\sum_{k=1}^nx_k)^{\frac{1}{\alpha}}(\sum_{k=1}^ny_k)^{\frac{1}{\beta}}$$

证明:由加权AG不等式或Young不等式,得:
$$\begin{aligned}(\frac{x_k^\alpha}{\sum_{i=1}^nx_i^\alpha})^{\frac{1}{\alpha}}(\frac{y_k^\beta}{\sum_{i=1}^ny_i^\beta})^{\frac{1}{\beta}}&\leq\frac{1}{\alpha}\cdot\frac{x_k^\alpha}{\sum_{i=1}^nx_i^\alpha}+\frac{1}{\beta}\cdot\frac{y_k^\beta}{\sum_{i=1}^ny_i^\beta}\\\sum_{i=1}^n\frac{x_iy_i}{(\sum_{i=1}^nx_i^\alpha)^\frac{1}{\alpha}(\sum_{i=1}^ny_i^\beta)^\frac{1}{\beta}}&\leq\frac{1}{\alpha}+\frac{1}{\beta}=1\\\therefore\quad\sum_{k=1}^nx_ky_k&\leq(\sum_{k=1}^nx_k)^{\frac{1}{\alpha}}(\sum_{k=1}^ny_k)^{\frac{1}{\beta}}\end{aligned}$$

而当取$\alpha=2,\beta=2$时,便可得“柯西不等式”

Schur不等式

假设$$1\leq{i,j,k}\leq{n}\\\sum_{i=1}^na_{ij}=1,\sum_{j=1}^na_{ij}=1\\x_k\geq{0},a_{ij}\geq{0}$$

则$$y_1y_2\cdots{y_n}\geq{x_1x_2\cdots{x_n}}$$

其中$y_i=\sum_{k=1}^na_{ik}x_k,(1\leq{i}\leq{n})$

证明:如果x中有一个为0,不等式自然成立。因此,考虑x正数的情况。由于函数$$f(x)-=\ln{x},(x>0)$$

为连续的凸函数,利用加权的Jensen不等式,有
$$\begin{aligned}-\sum_{j=1}^na_{ij}\ln{x_j}&=-\frac{\sum_{j=1}^na_{ij}\ln{x_j}}{\sum_{j=1}^na_{ij}}\\&\geq-\ln\frac{\sum_{j=1}^na_{ij}\ln{x_j}}{\sum_{j=1}^na_{ij}}\\&=-\ln{y_i}\\\therefore\quad{y_i}&\geq{x}_1^{a_{i1}}x_1^{a_{i2}}\cdots{x_1^{a_{in}}},(1\leq{i}\leq{n})\\\therefore\quad{y_1y_2\cdots{y_n}}&\geq(x_1^{a_{11}}x_1^{a_{12}}\cdots{x_1^{a_{1n}}})(x_1^{a_{21}}x_1^{a_{22}}\cdots{x_1^{a_{2n}}})\\&\quad\cdots(x_1^{a_{n1}}x_1^{a_{n2}}\cdots{x_1^{a_{nn}}})\\&=x_1^{\sum_{j=1}^na_{j1}}x_2^{\sum_{j=1}^na_{j2}}\cdots{x_n^{\sum_{j=1}^na_{jn}}}\\&=x_1x_2\cdots{x_n}\end{aligned}$$

霍尔德不等式

假设$$a_i\geq{0},b_i\geq{0},(1\leq{i}\leq{n})\\\alpha>0,\beta>0,\alpha+\beta=1\\$$

则$$\sum_{i=1}^na_i^\alpha{b_i^\beta}\leq(\sum_{i=1}^na_i)^\alpha(\sum_{j=1}^nb_j)^\beta$$

证明:令$A=\sum_{i=1}^na_i,B=\sum_{j=1}^nb_j$,那么
$$\begin{aligned}A^{-\alpha}B^{-\beta}\sum_{i=1}^na_i^{\alpha}b_i^{\beta}&=\sum_{i=1}^n(\frac{a_i}{A})^{\alpha}(\frac{b_i}{B})^{\beta}\\\alpha\ln\frac{a_i}{A}+\beta\ln\frac{b_i}{B}&=\frac{\alpha\ln\frac{a_i}{A}+\beta\ln\frac{b_i}{B}}{\alpha+\beta}\\&\leq\ln\frac{\alpha\frac{a_i}{A}+\beta\frac{b_i}{B}}{\alpha+\beta}\\&=\ln(\alpha\frac{a_i}{A}+\beta\frac{b_i}{B})\\\therefore\quad(\frac{a_i}{A})^{\alpha}(\frac{b_i}{B})^{\beta}&\leq\alpha\frac{a_i}{A}+\beta\frac{b_i}{B}\\\therefore\quad\sum_{i=1}^n(\frac{a_i}{A})^{\alpha}(\frac{b_i}{B})^{\beta}&\leq\frac{\alpha}{A}\sum_{i=1}^na_i+\frac{\beta}{B}\sum_{i=1}b_i\\&=\alpha+\beta\\&=1\\\therefore\quad\sum_{i=1}^na_i^\alpha{b_i^\beta}&\leq(\sum_{i=1}^na_i)^\alpha(\sum_{j=1}^nb_j)^\beta\end{aligned}$$

令$\alpha=\beta=\frac{1}{2},a_i=x_i^2,b_i=y_i^2$,带入“霍尔德不等式”可得“柯西不等式”

幂平均值不等式

若$a_i>0,(1\leq{i}\leq{n}),\alpha>\beta>0$则$$(\frac{1}{n}\sum_{i=1}^na_i^\alpha)^\frac{1}{\alpha}\geq(\frac{1}{n}\sum_{i=1}^na_i^\beta)^\frac{1}{\beta}$$

证明:使用霍尔德不等式,令
$$\begin{aligned}x_i&=1,(1\leq{i}\leq{n})\\\sum_{i=1}^nx_i^p{y_i^q}&\leq(\sum_{i=1}^nx_i)^p(\sum_{j=1}^ny_j)^q\\\therefore\quad\sum_{i=1}^ny_i&\leq{n}^\frac{1}{p}(\sum_{j=1}^ny_j^q)^\frac{1}{q}\\\because\quad\frac{1}{p}+\frac{1}{q}&=1,(q>1,p>1)\\\therefore\quad\frac{1}{n}\sum_{i=1}^ny_i&\leq(\frac{1}{n}\sum_{j=1}^ny_j^q)^\frac{1}{q}\\y_i&=a_i^\beta,q=\frac{\alpha}{\beta}>1\\\therefore\quad\frac{1}{n}\sum_{i=1}^na_i^\beta&\leq(\frac{1}{n}\sum_{j=1}^na_j^\alpha)^\frac{\beta}{\alpha}\\\therefore\quad(\frac{1}{n}\sum_{i=1}^na_i^\alpha)^\frac{1}{\alpha}&\geq(\frac{1}{n}\sum_{i=1}^na_i^\beta)^\frac{1}{\beta}\end{aligned}$$

Minkowski不等式

对$a_i\geq{0},b_i\geq{0},p>1$有$$(\sum_{i=1}^n(a_i+b_i)^p)^\frac{1}{p}\leq(\sum_{i=1}^na_i^p)^\frac{1}{p}+(\sum_{i=1}^nb_i^p)^\frac{1}{p}$$

证明:令$\frac{1}{p}+\frac{1}{q}=1$,由霍尔德不等式
$$\begin{aligned}\sum_{i=1}^na_i(a_i+b_i)^{p-1}&\leq(\sum_{i=1}^na_i^p)^\frac{1}{p}(\sum_{i=1}^n(a_i+b_i)^{(p-1)q})^\frac{1}{q}\\&=(\sum_{i=1}^na_i^p)^\frac{1}{p}(\sum_{i=1}^n(a_i+b_i)^p)^{1-\frac{1}{p}}\end{aligned}$$

$$\sum_{i=1}^nb_i(a_i+b_i)^{p-1}\leq(\sum_{i=1}^nb_i^p)^\frac{1}{p}(\sum_{i=1}^n(a_i+b_i)^p)^{1-\frac{1}{p}}\\\sum_{i=1}^n(a_i+b_i)^{p}\leq((\sum_{i=1}^na_i^p)^\frac{1}{p}+(\sum_{i=1}^nb_i^p)^\frac{1}{p})(\sum_{i=1}^n(a_i+b_i)^p)^{1-\frac{1}{p}}$$

$$\begin{aligned}\therefore\quad(\sum_{i=1}^n(a_i+b_i)^p)^\frac{1}{p}&\leq(\sum_{i=1}^na_i^p)^\frac{1}{p}+(\sum_{i=1}^nb_i^p)^\frac{1}{p}\\\end{aligned}$$

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