Cauchy-Schwarz不等式的证明与推广

Proof and Generalization of Cauchy-Schwarz Inequality

摘要: Cauchy-Schwarz不等式在数学领域中是一类重要的不等式。本文归纳了Cauchy-Schwarz不等式几种典型证明方法，并给出了其推广形式。

Abstract: Cauchy-schwarz inequality is an important inequality in mathematics. In this paper, several typical proof methods of Cauchy-Schwarz inequality are summarized and their generalized forms are given.

1. 引言

柯西不等式是一类著名不等式 [1] [2]，在实数域或复数域上的内积空间 $Y$， $\exists x,y\in Y$， ${\left|\left(x,y\right)\right|}^2\le {\Vert x\Vert }^2{\Vert y\Vert }^2$ 为Cauchy-Schwarz不等式最基本形式。此不等式在数学领域中有着广泛的应用，是区别于均值不等式的另一类重要不等式。通过变形可以推广出很多形式，从而应用在不同领域中。本文主要介绍了Cauchy-Schwarz不等式的离散形式与积分形式，并给出了几种具有代表性的证明方法，以便于对柯西不等式更好的理解。

2. Cauchy不等式的定理证明

2.1. 柯西不等式的离散形式

定理1 [3]：对 $\forall a_i,b_i\in R\left(i=1,2,\cdots ,n\right)$，有 ${\left({\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}\right)}^2\le \left({\displaystyle {\sum }_{i=1}^n{a}_i^2}\right)\left({\displaystyle {\sum }_{i=1}^n{b}_i^2}\right)$。等号成立的充要是， $\exists 实数k\in R$, $\text{st}{a}_i=k{b}_i$，即当 $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\dots \frac{a_n}{b_n}$ 时等号成立。

引理1 [2] [4] 设矩阵 $A=\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ \cdots & & \cdots \\ a_{m1}& \cdots & a_{mn}\end{array}\right)$, $B=\left(\begin{array}{ccc}{b}_{11}& \cdots & b_{1m}\\ \cdots & & \cdots \\ b_{n1}& \cdots & b_{nm}\end{array}\right)$，令 $C=AB=C_{ij}$, $C={\displaystyle {\sum }_{i,j=1}^n{a}_{ij}{b}_{ji}}$，则 $\left|C\right|$ 等于 $A$ 中所需要的 $m$ 阶子式与 $B$ 中对应同阶子式乘积和。

证法1 令矩阵 $A=\left(\begin{array}{cc}\begin{array}{ccc}{a}_1& a_2& \cdots \\ b_1& b_2& \cdots \end{array}& \begin{array}{c}{a}_n\\ b_n\end{array}\end{array}\right)$， $A^{\prime }=\left(\begin{array}{c}\begin{array}{cc}{a}_1& b_1\end{array}\\ \begin{array}{cc}{a}_2& b_2\\ \cdots & \cdots \\ a_n& b_n\end{array}\end{array}\right)$，则 $C=A{A}^{\prime }=\left(\begin{array}{cc}{\displaystyle {\sum }_{i=1}^n{a}_i^2}& {\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}\\ {\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}& {\displaystyle {\sum }_{i=1}^n{b}_i^2}\end{array}\right)$。

故 $\left|C\right|=\left({\displaystyle {\sum }_{i=1}^n{a}_i^2}\right)\left({\displaystyle {\sum }_{i=1}^n{b}_i^2}\right)-{\left({\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}\right)}^2$，由引理 $\left|C\right|={{\displaystyle \sum }}^{\text{}}\left|\begin{array}{cc}{a}_i& a_j\\ b_i& b_j\end{array}\right|\left|\begin{array}{cc}{a}_i& b_i\\ a_j& b_j\end{array}\right|={\displaystyle {\sum }_{1\le i\le j\le n}{\left(a_i{b}_j-a_j{b}_i\right)}^2}\ge 0$。

故 ${\left({\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}\right)}^2\le \left({\displaystyle {\sum }_{i=1}^n{a}_i^2}\right)\left({\displaystyle {\sum }_{i=1}^n{b}_i^2}\right)$。

证法2 [5] 构造二次函数

令

$\begin{array}{c}f\left(x\right)={\left(a_{1}x+b_1\right)}^2+{\left(a_{2}x+b_2\right)}^2+\cdots +{\left(a_{n}x+b_n\right)}^2\\ =\left(a_1^2+a_2^2+\cdots +a_n^2\right)x^2+2\left(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\right)x+\left(b_1^2+b_2^2+\cdots +b_n^2\right)\end{array}$

其中 $a_1^2+a_2^2+\cdots +a_n^2\ge 0$。

又 $\because \frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$ $\therefore f\left(x\right)\ge 0$

$\therefore \Delta =4{\left(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\right)}^2-4\left(a_1^2+a_2^2+\cdots +a_n^2\right)\left(b_1^2+b_2^2+\cdots +b_n^2\right)\le 0$ 恒成立

即 ${\left(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n\right)}^2\le \left(a_1^2+a_2^2+\cdots +a_n^2\right)\left(b_1^2+b_2^2+\cdots +b_n^2\right)$，当且仅当 $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$ 时等号成立。

引理2 在规定的欧式空间中，对 $\forall$ 两个向量，回顾两个性质：

1) 令 $\alpha =\left(x_1,x_2,\cdots ,x_n\right)$, $\beta =\left(y_1,y_2,\cdots ,y_n\right)$，有 $\left(\alpha ,\beta \right)=x_1{y}_1+x_2{y}_2+\cdots +x_n{y}_n$。

2) 在一个欧式空间里，对 $\forall$ 向量 $\alpha ,\beta$ 有不等式 ${\left(\alpha ,\beta \right)}^2\le \left(\alpha ,\alpha \right)\left(\beta ,\beta \right)$，等号成立当且仅当 $\alpha ,\beta$ 线性相关。

证法3 通过欧式空间，取 $\alpha =\left(x_1,x_2,\cdots ,x_n\right)$， $\beta =\left(y_1,y_2,\cdots ,y_n\right)$ 则 ${\left(\alpha ,\beta \right)}^2={\left(x_1{y}_1+x_2{y}_2+\cdots +x_n{y}_n\right)}^2$， ${\left(\alpha ,\alpha \right)}^2=x_1^2+x_2^2+\cdots +x_n^2$， ${\left(\beta ,\beta \right)}^2=y_1^2+y_2^2+\cdots +y_n^2$，根据引理 ${\left(\alpha ,\beta \right)}^2\le \left(\alpha ,\alpha \right)\left(\beta ,\beta \right)$ 可得 ${\left(x_1{y}_1+x_2{y}_2+\cdots +x_n{y}_n\right)}^2\le \left(x_1^2+x_2^2+\cdots +x_n^2\right)\left(y_1^2+y_2^2+\cdots +y_n^2\right)$ 等号成立当且仅当 $\alpha ,\beta$ 线性相关。

2.2. 柯西不等式的积分形式

定理2 [6] [7] (Cauchy-Schwarz不等式)设在 $\left[af\left(x\right)和g\left(x\right),b\right]$ 上的实可积函数，则 ${\left({\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2\le {\left({\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\right)}^2{\left({\displaystyle {\int }_a^{b}g\left(x\right)\text{d}x}\right)}^2$ (当且仅当 $f\left(x\right),g\left(x\right)$ 线性相关时等号成立)。

证法1 $\because f\left(x\right),g\left(x\right)$ 都在 $\left[a,b\right]$ 上可积，利用积分定义将 $\left[a,b\right]$ 区间 $n$ 等分

令 $x_i=a+\frac{i}{n}\left(b-a\right)\left(i=1,2,\cdots ,n\right)$，由定积分的性质 $f^2$, $f·g$，在 $\left[a,b\right]$ 上均可积

${\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}={\lim }_{n\to \infty }{\displaystyle {\sum }_{i=1}^{n}f\left(x_i\right)g\left(x_i\right)\frac{b-a}{n}}$

${\displaystyle {\int }_a^{b}f{\left(x\right)}^2\text{d}x}={\lim }_{n\to \infty }{\displaystyle {\sum }_{i=1}^{n}f{\left(x_i\right)}^2\frac{b-a}{n}}$

${\displaystyle {\int }_a^{b}g{\left(x\right)}^2\text{d}x}={\lim }_{n\to \infty }{\displaystyle {\sum }_{i=1}^{n}g{\left(x_i\right)}^2\frac{b-a}{n}}$

由Cauchy-Schwarz不等式离散形式 ${\left({\displaystyle {\sum }_{i=1}^n{a}_i{b}_i}\right)}^2\le \left({\displaystyle {\sum }_{i=1}^n{a}_i^2}\right)\left({\displaystyle {\sum }_{i=1}^n{b}_i^2}\right)$

得 ${\left({\displaystyle {\sum }_{i=1}^{n}f\left(x_i\right)g\left(x_i\right)}\right)}^2\le \left({\displaystyle {\sum }_{i=1}^{n}f{\left(x_i\right)}^2}\right)\left({\displaystyle {\sum }_{i=1}^{n}g{\left(x_i\right)}^2}\right)$ 并由极限的保号性证明成立即

${\left({\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2\le {\left({\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\right)}^2{\left({\displaystyle {\int }_a^{b}g\left(x\right)\text{d}x}\right)}^2$。

证法2(判别式法)

对 $\forall 实数m$，有 ${\displaystyle {\int }_a^b{\left(mf\left(x\right)+g\left(x\right)\right)}^2\text{d}x}\ge 0$

即

$m^2{\left({\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\right)}^2+2m{\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}+{\left({\displaystyle {\int }_a^{b}g\left(x\right)\text{d}x}\right)}^2\ge 0$

$\therefore \Delta ={\left(2{\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2-4{\left({\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\right)}^2{\left({\displaystyle {\int }_a^{b}g\left(x\right)\text{d}x}\right)}^2\le 0$

即 ${\left({\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2\le {\left({\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\right)}^2{\left({\displaystyle {\int }_a^{b}g\left(x\right)\text{d}x}\right)}^2$ 证明成立。

证法3 利用定积分的性质

令 $A\ge 0$，即 $A^2={\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}$

$B={\left[{\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}\right]}^{1/2}>0$ 即 $B^2={\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}$ 其中 $f\left(x\right)\cdot g\left(x\right)\ne 0$

$\left(\frac{\left|f\left(x\right)\right|}{A}-\frac{\left|g\left(x\right)\right|}{B}\right){\left[{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}\right]}^{1/2}{}^{{}^2}=\frac{f^2\left(x\right)}{A^2}+\frac{g^2\left(x\right)}{B^2}-\frac{2}{AB}\left|f\left(x\right)g\left(x\right)\right|$ 取积分

$\begin{array}{c}{\displaystyle {\int }_a^b{\left(\frac{\left|f\left(x\right)\right|}{A}-\frac{\left|g\left(x\right)\right|}{B}\right)}^2\text{d}x}={\displaystyle {\int }_a^b\frac{f^2\left(x\right)}{A^2}\text{d}x}+{\displaystyle {\int }_a^b\frac{g^2\left(x\right)}{B^2}\text{d}x}-\frac{2}{AB}{\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}\\ =\frac{{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}}{A^2}+\frac{{\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}}{B^2}-\frac{2}{AB}{\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}\\ =2-\frac{2}{AB}{\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}\ge 0\end{array}$

化简得 $\frac{{\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}}{AB}\le 1$ 因此 ${\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}\le AB$

由 $A={\left[{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}\right]}^{1/2}$, $B={\left[{\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}\right]}^{1/2}$

代入得 ${\displaystyle {\int }_a^b\left|f\left(x\right)g\left(x\right)\right|\text{d}x}\le {\left[{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}\right]}^{1/2}\cdot {\left[{\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}\right]}^{1/2}$

因此 ${\left({\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2\le \left({\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}\right)\left({\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}\right)$。

3. Cauchy-Schwarz不等式其他推广及应用

定理3 [8] 将不等式 ${\left({\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\right)}^2\le \left({\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}\right)\left({\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}\right)$ 改写成行列式的形式 $\left|\begin{array}{cc}{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}g\left(x\right)f\left(x\right)\text{d}x}\\ {\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}& {\displaystyle {\int }_a^b{g}^2(x)\text{d}x}\end{array}\right|>0$，再设另一函数 $u\left(x\right)$， $\text{st}f\left(x\right),g\left(x\right),u\left(x\right)$ 在 $\left[a,b\right]$ 上可积，那么 $\left|\begin{array}{ccc}{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}g\left(x\right)f\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}u\left(x\right)f\left(x\right)\text{d}x}\\ {\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}}x& {\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}u\left(x\right)g\left(x\right)\text{d}x}\\ {\displaystyle {\int }_a^{b}f\left(x\right)u\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}g\left(x\right)u\left(x\right)\text{d}x}& {\displaystyle {\int }_a^b{u}^2\left(x\right)\text{d}x}\end{array}\right|\ge 0$。

证明 对 $\forall 实数m_1,m_2,m_3$，有

$\begin{array}{l}{\displaystyle {\int }_a^b{\left(m_{1}f\left(x\right)+m_{2}g\left(x\right)+m_{3}u\left(x\right)\right)}^2\text{d}x}\\ =m_1^2{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x+m_2^2\underset{a}{\overset{b}{{\displaystyle \int }}}{g}^2\left(x\right)\text{d}x}+m_3^2{\displaystyle {\int }_a^b{u}^2\left(x\right)\text{d}x}+2m_1{m}_2{\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}\\ +2m_1{m}_3{\displaystyle {\int }_a^{b}f\left(x\right)u\left(x\right)\text{d}x}+2m_2{m}_3{\displaystyle {\int }_a^{b}g\left(x\right)u\left(x\right)\text{d}x}\ge 0\end{array}$

可以看到 $m_1,m_2,m_3$ 的二次型为半正定二次型，从而系数矩阵行列式为

$\left|\begin{array}{ccc}{\displaystyle {\int }_a^b{f}^2\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}g\left(x\right)f\left(x\right)\text{d}}x& {\displaystyle {\int }_a^{b}u\left(x\right)f\left(x\right)\text{d}x}\\ {\displaystyle {\int }_a^{b}f\left(x\right)g\left(x\right)\text{d}x}& {\displaystyle {\int }_a^b{g}^2\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}u\left(x\right)g\left(x\right)\text{d}x}\\ {\displaystyle {\int }_a^{b}f\left(x\right)u\left(x\right)\text{d}x}& {\displaystyle {\int }_a^{b}g\left(x\right)u\left(x\right)\text{d}x}& {\displaystyle {\int }_a^b{u}^2\left(x\right)\text{d}x}\end{array}\right|\ge 0$。

引理3 [9] [10] $\forall \alpha ,\beta \ge 0$。 $\alpha ,\beta \in R$。有 $\alpha \beta \le \frac{{\alpha }^p}{p}+\frac{{\beta }^q}{q}$， $\left(p,q>1\right)$，且 $\frac{1}{p}+\frac{1}{q}=1$。

定理4

1) 在欧式空间 $R^n$，若 $\xi$， $\eta \in {\text{R}}^{\text{n}}$，则 ${\left(\xi ,\eta \right)}^2\le \left(\xi ,\xi \right)\left(\eta ,\eta \right)$，当且仅当线性相关时等号成立。

2) 设概率空间 $P$ 中， $\forall$ 随机变量 $\xi$， $\eta \in P$。有 ${\left|E\xi \eta \right|}^2\le {\left|E\xi \right|}^2{\left|E\eta \right|}^2$，当且仅当 $\eta =C\xi$ 时，等号成立， $C$ 为任意常数。

证明 1) 设 $\eta \ne 0$。令 $m$ 是一个实数，作向量 $\gamma =\xi +m\eta$，不论 $m$ 取何值一定有

$\left(\gamma ,\gamma \right)=\left(\xi +m\eta ,\xi +m\eta \right)\ge 0$

即 $\left(\xi ,\xi \right)+2\left(\xi ,\eta \right)m+\left(\eta ,\eta \right)m^2\ge 0$

取 $m=-\frac{\left(\xi ,\eta \right)}{\left(\eta ,\eta \right)}$ 代入(1)式得

$\left(\xi ,\xi \right)-\frac{{\left(\xi ,\eta \right)}^2}{\left(\eta ,\eta \right)}\ge 0$

${\left(\xi ,\eta \right)}^2\le \left(\xi ,\xi \right)\left(\eta ,\eta \right)$

证明成立。

2) 当 $E\left({\xi }^2\right)=0$ 时， $\xi =0$，故 $E\left(\xi \eta \right)=0$ 不等式成立；当 $E\left({\xi }^2\right)>0$ 时，令 $f\left(t\right)=E\left[{\left(t\xi -\eta \right)}^2\right]=t^{2}E\left({\xi }^2\right)-2tE\left(\xi ,\eta \right)+E\left({\eta }^2\right)\ge 0$，其中 $f\left(t\right)$ 大于等于0，于是 $\Delta =4{\left(E\xi \eta \right)}^2-4E\left({\xi }^2\right)E\left({\eta }^2\right)\le 0$，即 ${\left(E\xi \eta \right)}^2\le E\left({\xi }^2\right)E\left({\eta }^2\right)$，证明成立。

定理5 [11] 对 $\forall {\xi }_k$， ${\eta }_k\in C\left(k=1,2,\cdots ,n\right)$，有 ${\displaystyle {\sum }_{i=1}^n\left|{\xi }_k\right|\left|{\eta }_k\right|}\le {\left({\displaystyle {\sum }_{k=1}^n{\left|{\xi }_k\right|}^p}\right)}^{1/p}\cdot {\left({\displaystyle {\sum }_{k=1}^n{\left|{\eta }_k\right|}^q}\right)}^{1/q}$， $\left(p>1,q>1且\frac{1}{p}+\frac{1}{q}=1\right)$。

证明 当 $\xi =\eta =0\left(k=1,2,\cdots ,n\right)$ 时，结论成立。设 $\xi$ 不全为0， $\eta$ 也不全为零，由引理3，得

$\begin{array}{c}\frac{{\displaystyle {\sum }_{k=1}^n\left|\xi \right|\left|\eta \right|}}{{\left({\displaystyle {\sum }_{k=1}^n{\left|\xi \right|}^p}\right)}^{1/p}{\left({\displaystyle {\sum }_{k=1}^n{\left|\eta \right|}^q}\right)}^{1/q}}={\displaystyle {\sum }_{k=1}^n\left[\frac{\left|\xi \right|}{{\left({\displaystyle {\sum }_{k=1}^n{\left|\xi \right|}^p}\right)}^{1/p}}\right]\left[\frac{\left|\eta \right|}{{\left({\displaystyle {\sum }_{k=1}^n{\left|\eta \right|}^q}\right)}^{1/q}}\right]}\\ \le {\displaystyle {\sum }_{k=1}^n\left[\frac{{\left|\xi \right|}^p}{p\left({\displaystyle {\sum }_{k=1}^n{\left|\xi \right|}^p}\right)}+\frac{{\left|\eta \right|}^q}{q\left({\displaystyle {\sum }_{k=1}^n{\left|\eta \right|}^q}\right)}\right]}\\ =\frac{1}{p}+\frac{1}{q}=1.\end{array}$

因此结论成立。