# 高中数学/不等式与数列/柯西不等式

## 基础知识

### 柯西不等式的定义与证明

${\displaystyle \forall {\vec {a}},{\vec {b}}\in \mathbb {R} ^{n},s.t.|{\vec {a}}||{\vec {b}}|\geq {\vec {a}}\cdot {\vec {b}}}$

${\displaystyle {\begin{array}{l}\forall a_{i},b_{i}\in \mathbb {R} ,i=1,2,...,n,\\s.t.(a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2})(b_{1}^{2}+b_{2}^{2}+\cdots +b_{n}^{2})\geq (a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n})^{2}\end{array}}}$

### 无特殊条件约束的简单应用

相关例题1： 求${\displaystyle (a^{2}+b^{2}+c^{2})({\frac {1}{a^{2}}}+{\frac {4}{b^{2}}}+{\frac {9}{c^{2}}})}$ 的最小值。

${\displaystyle {\begin{array}{l}(a^{2}+b^{2}+c^{2})({\frac {1}{a^{2}}}+{\frac {4}{b^{2}}}+{\frac {9}{c^{2}}})=(a^{2}+b^{2}+c^{2})(({\frac {1}{a}})^{2}+({\frac {2}{b}})^{2}+({\frac {3}{c}})^{2})\\\geq (a\cdot {\frac {1}{a}}+b\cdot {\frac {2}{b}}+c\cdot {\frac {3}{c}})^{2}=(1+2+3)^{2}=6^{2}=36\end{array}}}$

相关例题2： 设${\displaystyle a,b,c>0}$ ，求${\displaystyle (a+b+c)({\frac {1}{a}}+{\frac {4}{b}}+{\frac {1}{c}})}$ 的最小值。

相关例题3： 设${\displaystyle a,b,c>0}$ ，求${\displaystyle (a+b^{2}+c^{3})({\frac {8}{a}}+{\frac {4}{b^{2}}}+{\frac {2}{c^{3}}})}$ 的最小值。

相关例题4： 求${\displaystyle (a^{2}+b^{-2}+2)(b^{2}+a^{-2}+2)}$ 的最小值。

相关例题5： 设${\displaystyle 0 ，求函数${\displaystyle f(x)={\frac {2}{x}}+{\frac {9}{1-2x}}}$ 的最小值。

${\displaystyle {\begin{array}{l}f(x)=({\frac {2}{x}}+{\frac {9}{1-2x}})\times 1=({\frac {4}{2x}}+{\frac {9}{1-2x}})((2x)+(1-2x))\\=(({\frac {2}{\sqrt {2x}}})^{2}+({\frac {3}{\sqrt {1-2x}}})^{2})(({\sqrt {2x}})^{2}+({\sqrt {1-2x}})^{2})\\\geq ({\frac {2}{\sqrt {2x}}}\cdot {\sqrt {2x}}+{\frac {3}{\sqrt {1-2x}}}\cdot {\sqrt {1-2x}})^{2}\\=(2+3)^{2}=5^{2}=25\end{array}}}$

相关例题6： 求函数${\displaystyle f(x)=5{\sqrt {x-1}}+{\sqrt {10-2x}}}$ 的最大值。

${\displaystyle \left\{{\begin{array}{l}x-1\geq 0\\10-2x\geq 0\end{array}}\right.}$

${\displaystyle {\begin{array}{l}(f(x))^{2}=(5\times {\sqrt {x-1}}+{\sqrt {2}}\times {\sqrt {5-x}})^{2}\\\leq (5^{2}+({\sqrt {2}})^{2})(({\sqrt {x-1}})^{2}+({\sqrt {5-x}})^{2})=27((x-1)+(5-x))=27\times 4=108\\\Rightarrow f(x)\leq {\sqrt {108}}=6{\sqrt {3}}\end{array}}}$

相关例题7： 求证：${\displaystyle (a^{4}+b^{4})(a^{2}+b^{2})\geq (a^{3}+b^{3})^{2}}$

相关例题8： 设${\displaystyle m,n>0}$ ，求证：${\displaystyle {\frac {a^{2}}{m}}+{\frac {b^{2}}{n}}\geq {\frac {(a+b)^{2}}{m+n}}}$

${\displaystyle {\begin{array}{l}{\frac {a^{2}}{m}}+{\frac {b^{2}}{n}}\geq {\frac {(a+b)^{2}}{m+n}}\\\Leftrightarrow ({\frac {a^{2}}{m}}+{\frac {b^{2}}{n}})(m+n)\geq (a+b)^{2}\\\Leftrightarrow (({\frac {a}{\sqrt {m}}})^{2}+({\frac {b}{\sqrt {n}}})^{2})(({\sqrt {m}})^{2}+({\sqrt {n}})^{2})\geq ({\frac {a}{\sqrt {m}}}\cdot {\sqrt {m}}+{\frac {b}{\sqrt {n}}}\cdot {\sqrt {n}})^{2}\end{array}}}$

### 比较直接的条件代换

相关例题1： 设${\displaystyle a,b,c>0,a+b+c=1}$ ，求证：${\displaystyle a^{2}+b^{2}+c^{2}\geq {\frac {1}{3}}}$

${\displaystyle {\begin{array}{l}(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2})\geq (a+b+c)^{2}=1\\\Rightarrow a^{2}+b^{2}+c^{2}\geq {\frac {1}{3}}\end{array}}}$

相关例题2： 设${\displaystyle a,b,c>0,a+b+c=7}$ ，求证：${\displaystyle a^{2}+4b^{2}+9c^{2}\geq 36}$

${\displaystyle {\begin{array}{l}(a^{2}+4b^{2}+9c^{2})(1+{\frac {1}{4}}+{\frac {1}{9}})\\=(a^{2}+(2b)^{2}+(3c)^{2})(1^{2}+({\frac {1}{2}})^{2}+({\frac {1}{3}})^{2})\\\geq (a\times 1+2b\times {\frac {1}{2}}+3c\times {\frac {1}{3}})^{2}\\\Rightarrow (a^{2}+4b^{2}+9c^{2})\times {\frac {49}{36}}\geq (a+b+c)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}{\frac {49}{36}}(a^{2}+4b^{2}+9c^{2})\geq 7^{2}=49\\\Rightarrow a^{2}+4b^{2}+9c^{2}\geq 49\times {\frac {36}{49}}=36\end{array}}}$

相关例题3： 设${\displaystyle a,b,c>0,a+2b+3c=6}$ ，求证：${\displaystyle a^{2}+b^{2}+c^{2}\geq {\frac {18}{7}}}$

${\displaystyle {\begin{array}{l}(a^{2}+b^{2}+c^{2})(1+4+9)\\=(a^{2}+b^{2}+c^{2})(1^{2}+2^{2}+3^{2})\\\geq (a\times 1+b\times 2+c\times 3)^{2}\\\Rightarrow (a^{2}+b^{2}+c^{2})\times 14\geq (a+2b+3c)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}14(a^{2}+b^{2}+c^{2})\geq 6^{2}=36\\\Rightarrow a^{2}+b^{2}+c^{2}\geq {\frac {18}{7}}\end{array}}}$

相关例题4： 设${\displaystyle a,b,c>0,a+2b+3c=6}$ ，求证：${\displaystyle a^{2}+2b^{2}+3c^{2}\geq 6}$

${\displaystyle {\begin{array}{l}(a^{2}+2b^{2}+3c^{2})(1+2+3)\\=(a^{2}+({\sqrt {2}}b)^{2}+({\sqrt {3}}c)^{2})(({\sqrt {1}})^{2}+({\sqrt {2}})^{2}+({\sqrt {3}})^{2})\\\geq (a\times {\sqrt {1}}+({\sqrt {2}}b)\times {\sqrt {2}}+({\sqrt {3}}c)\times {\sqrt {3}})^{2}\\\Rightarrow (a^{2}+2b^{2}+3c^{2})\times 6\geq (a+2b+3c)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}6(a^{2}+2b^{2}+3c^{2})\geq 6^{2}\\\Rightarrow a^{2}+2b^{2}+3c^{2}\geq 6\end{array}}}$

相关例题5： 设${\displaystyle a,b,c>0,{\frac {(a-1)^{2}}{16}}+{\frac {(b+2)^{2}}{5}}+{\frac {(c-3)^{2}}{4}}=1}$ ，求${\displaystyle a+b+c}$ 的最大值和最小值。

${\displaystyle {\begin{array}{l}({\frac {(a-1)^{2}}{16}}+{\frac {(b+2)^{2}}{5}}+{\frac {(c-3)^{2}}{4}})(4^{2}+({\sqrt {5}})^{2}+2^{2})\geq ({\frac {a-1}{4}}\times 4+{\frac {b+2}{\sqrt {5}}}\times {\sqrt {5}}+{\frac {c-3}{2}}\times 2)^{2}\\\Rightarrow ({\frac {(a-1)^{2}}{16}}+{\frac {(b+2)^{2}}{5}}+{\frac {(c-3)^{2}}{4}})(16+5+4)\geq ((a-1)+(b+2)+(c-3))^{2}\\\Rightarrow ({\frac {(a-1)^{2}}{16}}+{\frac {(b+2)^{2}}{5}}+{\frac {(c-3)^{2}}{4}})\times 25\geq ((a+b+c)-1+2-3)^{2}=((a+b+c)-2)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}1\times 25\geq ((a+b+c)-2)^{2}\\\Rightarrow -5\leq (a+b+c)-2\leq 5\\\Rightarrow -3\leq (a+b+c)\leq 7\end{array}}}$

相关例题6： 设${\displaystyle x,y,z\in \mathbb {R} ,x^{2}+y^{2}+z^{2}=4}$ ，求${\displaystyle x-2y+2z}$ 的最小值。

${\displaystyle {\begin{array}{l}(x^{2}+y^{2}+z^{2})(1^{2}+(-2)^{2}+2^{2})\geq (x-2y+2z)^{2}\\\Rightarrow (x^{2}+y^{2}+z^{2})(1+4+4)\geq (x-2y+2z)^{2}\\\Rightarrow (x^{2}+y^{2}+z^{2})\times 9\geq (x-2y+2z)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}4\times 9\geq (x-2y+2z)^{2}\\\Rightarrow (x-2y+2z)^{2}\leq 36=6^{2}\\\Rightarrow -6\leq (x-2y+2z)^{2}\leq 6\end{array}}}$

相关例题7： 设${\displaystyle a,b,c\in \mathbb {R} ,2a-3b+c=3}$ ，求${\displaystyle a^{2}+(b-1)^{2}+c^{2}}$ 的最小值。

${\displaystyle {\begin{array}{l}(a^{2}+(b-1)^{2}+c^{2})(2^{2}+(-3)^{2}+1^{2})\geq (2a-3(b-1)+c)^{2}\\\Rightarrow (a^{2}+(b-1)^{2}+c^{2})(4+9+1)\geq (2a-3(b-1)+c)^{2}\\\Rightarrow (a^{2}+(b-1)^{2}+c^{2})\times 14\geq ((2a-3b+c)+3)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}(a^{2}+(b-1)^{2}+c^{2})\times 14\geq (3+3)^{2}\\\Rightarrow (a^{2}+(b-1)^{2}+c^{2})\geq {\frac {36}{14}}={\frac {18}{7}}\end{array}}}$

相关例题8： 设${\displaystyle a,b,c>0,a+b+c=1}$ ，求证：${\displaystyle {\frac {1}{a}}+{\frac {4}{b}}+{\frac {9}{c}}\geq 36}$

${\displaystyle {\begin{array}{l}({\frac {1}{a}}+{\frac {4}{b}}+{\frac {9}{c}})(a+b+c)\\=(({\frac {1}{\sqrt {a}}})^{2}+({\frac {2}{\sqrt {b}}})^{2}+({\frac {3}{\sqrt {c}}})^{2})(({\sqrt {a}})^{2}+({\sqrt {b}})^{2}+({\sqrt {c}})^{2})\\\geq ({\frac {1}{\sqrt {a}}}\times {\sqrt {a}}+{\frac {2}{\sqrt {b}}}\times {\sqrt {b}}+{\frac {3}{\sqrt {c}}}\times {\sqrt {c}})^{2}\\=(1+2+3)^{2}=36\end{array}}}$

${\displaystyle ({\frac {1}{a}}+{\frac {4}{b}}+{\frac {9}{c}})(a+b+c)\geq 36}$

相关例题9： 设${\displaystyle a,b>0,a+b=2}$ ，求证：${\displaystyle {\frac {a^{2}}{2-a}}+{\frac {b^{2}}{2-b}}\geq 2}$

${\displaystyle {\begin{array}{l}(({\frac {a}{\sqrt {2-a}}})^{2}+({\frac {b}{\sqrt {2-b}}})^{2})(({\sqrt {2-a}})^{2}+({\sqrt {2-b}})^{2})\geq (a+b)^{2}\\\Rightarrow ({\frac {a^{2}}{2-a}}+{\frac {b^{2}}{2-b}})((2-a)+(2-b))\geq (a+b)^{2}\\\Rightarrow ({\frac {a^{2}}{2-a}}+{\frac {b^{2}}{2-b}})(4-(a+b))\geq (a+b)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}({\frac {a^{2}}{2-a}}+{\frac {b^{2}}{2-b}})(4-2)\geq 2^{2}\\\Rightarrow {\frac {a^{2}}{2-a}}+{\frac {b^{2}}{2-b}}\geq 2\end{array}}}$

相关例题10： 设${\displaystyle 2a+3b+5c=29}$ ，求${\displaystyle {\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}}}$ 的最大值。

${\displaystyle {\begin{array}{l}(({\sqrt {2a+1}})^{2}+({\sqrt {3b+4}})^{2}+({\sqrt {5c+6}})^{2})(1^{2}+1^{2}+1^{2})\geq ({\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}})^{2}\\\Rightarrow ((2a+1)+(3b+4)+(5c+6))\times 3\geq ({\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}})^{2}\\\Rightarrow ((2a+3b+5c)+11)\times 3\geq ({\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}})^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}(29+11)\times 3\geq ({\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}})^{2}\\\Rightarrow {\sqrt {2a+1}}+{\sqrt {3b+4}}+{\sqrt {5c+6}}\leq {\sqrt {(29+11)\times 3}}={\sqrt {120}}=2{\sqrt {30}}\end{array}}}$

## 补充习题

• 已知${\displaystyle a,b,c>0}$ ，请分别使用平均值不等式和柯西不等式证明：${\displaystyle a+b+c\geq {\sqrt {ab}}+{\sqrt {bc}}+{\sqrt {ca}}}$
• ${\displaystyle a,b,c>0,a+b+c=1}$ ，求${\displaystyle a^{2}+b^{2}+c^{2}}$ 的最小值。
• ${\displaystyle a,b\in \mathbb {R} ,a^{2}+b^{2}=10}$ ，求${\displaystyle 3a+b}$ 的最大值和最小值。
• 已知${\displaystyle a,b\in \mathbb {R} ,3x^{2}+2y^{2}\leq 6}$ ，求${\displaystyle 2x+y}$ 的最大值和最小值。
• ${\displaystyle a,b,c>0,a+b+c=9}$ ，求${\displaystyle {\frac {4}{a}}+{\frac {9}{b}}+{\frac {16}{c}}}$ 的最小值。
（答案：9。）
• ${\displaystyle a,b,c>0,a+2b+3c=2}$ ，求${\displaystyle {\frac {1}{a}}+{\frac {2}{b}}+{\frac {3}{c}}}$ 的最小值。
（答案：18。）
• ${\displaystyle a,b,c>0}$ ，求证：${\displaystyle {\frac {2}{a+b}}+{\frac {2}{b+c}}+{\frac {2}{c+a}}\geq {\frac {9}{a+b+c}}}$
（提示：${\displaystyle 2(a+b+c)=(a+b)+(b+c)+(c+a)}$ 。）
• 已知函数${\displaystyle f(x)={\frac {1}{x^{2}}}+{\frac {4}{1-x^{2}}}\quad (-1
(1)求${\displaystyle f(x)}$ 的最小值。
(2)若${\displaystyle |t+1|\leq f(x)}$ 恒成立，求t的取值范围。

(1)首先由柯西不等式可知：
${\displaystyle {\begin{array}{l}f(x)={\frac {1}{x^{2}}}+{\frac {4}{1-x^{2}}}=({\frac {1}{x^{2}}}+{\frac {4}{1-x^{2}}})(x^{2}+(1-x^{2}))\\\geq (1+2)^{2}=9\\\Rightarrow f(x)\geq 9\end{array}}}$

(2) 因为${\displaystyle f(x)\geq 3}$ ，所以要使${\displaystyle |t+1|\leq f(x)}$ 恒成立，只需要使${\displaystyle |t+1|}$ 不超过${\displaystyle f(x)}$ 的最小值即可。
${\displaystyle {\begin{array}{l}|t+1|\leq min\{f(x)\}=3\\\Rightarrow -3\leq |t+1|\leq 3\end{array}}}$

## 参考资料

1. 俞求是; 章建跃; 田载今; 马波; 李世杰. 第3讲“柯西不等式”第1节“二维形式的柯西不等式”和第2节“一般形式的柯西不等式”. (编) 刘绍学 (主编); 钱珮玲 (副主编); 李龙才 (责任编辑). 高中数学 (A版) 选修4-5 2. 中国北京市海淀区中关村南大街17号院1号楼: 人民教育出版社. 2007: 31–41. ISBN 978-7-107-18675-2 （中文（中国大陆））.