# 高中数学/不等式与数列/常用不等式补充

## 基础知识

### n元的代数-几何平均值不等式

#### 代数-几何平均值不等式的一般形式及其证明

${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$${\displaystyle n}$个正实数，称${\displaystyle \mathbf {A} _{n}={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}}$为它们的算术平均数arithmetic mean），称${\displaystyle \mathbf {G} _{n}={\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}}$为它们的几何平均数geometric mean）。

${\displaystyle \mathbf {A} _{n}\geq \mathbf {G} _{n}}$

#### 3元情形下的应用

${\displaystyle {\begin{array}{l}f(x)=-x^{2}(x+1)+(x+1)=(1-x^{2})(x+1)=(1-x)(1+x)^{2}\\=4\cdot (1-x)\cdot {\frac {1+x}{2}}\cdot {\frac {1+x}{2}}\leq 4((1-x)+{\frac {{\frac {x+1}{2}}+{\frac {x+1}{2}}}{3}})^{3}={\frac {32}{27}}\end{array}}}$

${\displaystyle f(x)={\sqrt {x^{4}(1-x^{2})}}={\sqrt {4\cdot {\frac {x^{2}}{2}}\cdot {\frac {x^{2}}{2}}\cdot (1-x^{2})}}}$

${\displaystyle {\frac {x^{2}}{2}}\cdot {\frac {x^{2}}{2}}\cdot (1-x^{2})\leq ({\frac {{\frac {x^{2}}{2}}+{\frac {x^{2}}{2}}+(1-x^{2})}{3}})^{3}={\frac {1}{27}}}$

${\displaystyle f(x)\leq ({\frac {2x+(2+x)+3(2-x)}{3}})^{3}={\frac {8^{3}}{27}}}$

${\displaystyle {\begin{array}{l}f(x)^{2}=(6x(4-x^{2}))^{2}=36x^{2}(4-x^{2})^{2}\\=18\cdot 2x^{2}\cdot (4-x^{2})\cdot (4-x^{2})\\\leq 18\cdot ({\frac {2x^{2}+(4-x^{2})+(4-x^{2})}{3}})^{3}=18\cdot {\frac {8^{3}}{27}}\end{array}}}$

• 待求证或求解的式子经常可以拆成多个形式等价的部分，先用不等式单独求解。
• 合并单独的不等式时，需要保证取等条件一致。为此，所需待定常数的取值一般会结合原不等式的取等条件作为线索来猜测。

${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}a^{3}+1^{3}+1^{3}\geq 3{\sqrt[{3}]{a^{3}\cdot 1^{3}\cdot 1^{3}}}=3a\\b^{3}+1^{3}+1^{3}\geq 3{\sqrt[{3}]{a^{3}\cdot 1^{3}\cdot 1^{3}}}=3b\end{array}}\right.\\\Rightarrow a^{3}+b^{3}+4=(a^{3}+1^{3}+1^{3})+(b^{3}+1^{3}+1^{3})\geq 3a+3b=3(a+b)\\\Rightarrow {\frac {a^{3}+b^{3}+4}{3}}\geq a+b\quad \Rightarrow \quad {\frac {2+4}{3}}\geq a+b\quad \Rightarrow \quad a+b\leq 2\end{array}}}$

${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}1\cdot a\cdot a\leq {\frac {1^{3}+a^{3}+a^{3}}{3}}\\1\cdot b\cdot b\leq {\frac {1^{3}+b^{3}+b^{3}}{3}}\\1\cdot a\cdot b\leq {\frac {1^{3}+a^{3}+b^{3}}{3}}\end{array}}\right.\\\Rightarrow a^{2}+b^{2}+5ab\leq {\frac {1+a^{3}+a^{3}}{3}}+{\frac {1+b^{3}+b^{3}}{3}}+5\cdot {\frac {1+a^{3}+b^{3}}{3}}={\frac {7+7(x^{3}+y^{3})}{3}}\end{array}}}$

${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}a^{3}+b^{3}+1^{3}\geq 3\cdot a\cdot b\cdot 1\\b^{3}+c^{3}+1^{3}\geq 3\cdot b\cdot c\cdot 1\\c^{3}+a^{3}+1^{3}\geq 3\cdot c\cdot a\cdot 1\end{array}}\right.\\\Rightarrow 3+2(a^{3}+b^{3}+c^{3})=(a^{3}+b^{3}+1^{3})+(b^{3}+c^{3}+1^{3})+(b^{3}+a^{3}+1^{3})\\\geq (3\cdot a\cdot b\cdot 1)+3(\cdot b\cdot c\cdot 1)+3(\cdot c\cdot a\cdot 1)\\=3(ab+bc+ca)=2(ab+bc+ca)+(ab+bc+ca)\geq 2(ab+bc+ca)+3{\sqrt[{3}]{ab\cdot bc\cdot ca}}\\=2(ab+bc+ca)+3{\sqrt[{3}]{(abc)^{2}}}=3{\sqrt[{3}]{1^{2}}}\end{array}}}$

${\displaystyle a^{3}+b^{3}+c^{3}\geq ab+bc+ca}$。证明完毕。

${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}a^{3}+1^{3}+1^{3}\geq 3(1\cdot 1\cdot a)=3a\\b^{3}+1^{3}+1^{3}\geq 3(1\cdot 1\cdot b)=3b\end{array}}\right.\Rightarrow a^{3}+b^{3}+4=(a^{3}+2)+(b^{3}+2)\geq 3a+3b=3(a+b)\end{array}}}$

${\displaystyle a^{3}+b^{3}\geq 3(a+b)-4=3\cdot 2-4=2}$

${\displaystyle a^{2}+b^{2}+c^{2}+2ab+2bc+2ac=1}$

${\displaystyle (a^{2}+b^{2})+(b^{2}+c^{2})+(c^{2}+a^{2})+4(ab+bc+ac)=2}$

${\displaystyle {\sqrt {\frac {a^{2}+b^{2}+c^{2}}{3}}}\geq {\frac {a+b+c}{3}}={\frac {1}{3}}}$
${\displaystyle a^{2}+b^{2}+c^{2}}$的最小值为${\displaystyle {\frac {1}{3}}}$

${\displaystyle ab+bc+ca}$取最大值时，${\displaystyle a^{2}+b^{2}+c^{2}}$恰好取最小值。
${\displaystyle ab+bc+ca}$的最大值也为${\displaystyle {\frac {1}{3}}}$

(1)${\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\geq 3{\sqrt {3}}}$
(2)${\displaystyle abc(a+b+c)\leq {\frac {1}{3}}}$

(1)${\displaystyle ab+bc+ca\geq 3{\sqrt[{3}]{(ab)(bc)(ca)}}=3(abc)^{\frac {2}{3}}}$

${\displaystyle {\begin{array}{l}1\geq 3(abc)^{\frac {2}{3}}\quad \Rightarrow \quad abc\leq {\frac {1}{3{\sqrt {3}}}}\\\Rightarrow {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}={\frac {ab+bc+ac}{abc}}={\frac {1}{abc}}\leq 3{\sqrt {3}}\end{array}}}$

(2)首先注意到下列2个条件：
${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}abc(a+b+c)=(ab)(ca)+(bc)(ab)+(ca)(bc)\\ab+bc+ca=1\end{array}}\right.\end{array}}}$

${\displaystyle x=ab>0,y=bc>0,z=ca>0}$

${\displaystyle {\begin{array}{l}2(x+y+z)^{2}=(x^{2}+y^{2})+(y^{2}+z^{2})+(z^{2}+x^{2})+4(xy+yz+zx)\\\geq 2xy+2yz+2zx+4(xy+yz+zx)=6(xy+yz+zx)\end{array}}}$

${\displaystyle {\begin{array}{l}abc(a+b+c)=(ab)(ca)+(bc)(ab)+(ca)(bc)\\=xy+yz+zx\leq {\frac {1}{3}}(x+y+z)^{2}={\frac {1}{3}}\end{array}}}$

${\displaystyle x=a+{\frac {1}{a}}\geq 2{\sqrt {a{\frac {1}{a}}}}=2}$

${\displaystyle {\begin{array}{l}{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}-{\sqrt {2}}\geq a+{\frac {1}{a}}-2\\\Leftrightarrow {\sqrt {(a+{\frac {1}{a}})^{2}-2}}-{\sqrt {2}}\geq (a+{\frac {1}{a}})-2\\\Leftrightarrow {\sqrt {x^{2}-2}}-{\sqrt {2}}\geq x-2\\\Leftrightarrow {\sqrt {x^{2}-2}}\geq x+{\sqrt {2}}-2>0\\\Leftrightarrow ({\sqrt {x^{2}-2}})^{2}\geq (x+({\sqrt {2}}-2))^{2}\\\Leftrightarrow x^{2}-2\geq x^{2}+2({\sqrt {2}}-2)x+({\sqrt {2}}-2)^{2}\\\Leftrightarrow x^{2}-2\geq x^{2}+2({\sqrt {2}}-2)x+(2-4{\sqrt {2}}+4)\\\Leftrightarrow 4{\sqrt {2}}-8\geq 2({\sqrt {2}}-2)x\quad \Leftrightarrow \quad 2\geq x\end{array}}}$

${\displaystyle a^{2}+{\frac {1}{a^{2}}}\geq 2}$

${\displaystyle {\begin{array}{l}\Rightarrow {\sqrt {a^{2}+{\frac {1}{a^{2}}}}}\geq {\sqrt {2}}\\\Rightarrow (a+{\frac {1}{a}})+{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}\geq 2+{\sqrt {2}}\\\Rightarrow {\frac {1}{(a+{\frac {1}{a}})+{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}}}\leq {\frac {1}{2+{\sqrt {2}}}}\\\end{array}}}$

${\displaystyle {\begin{array}{l}{\frac {(a+{\frac {1}{a}})-{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}}{((a+{\frac {1}{a}})+{\sqrt {a^{2}+{\frac {1}{a^{2}}}}})((a+{\frac {1}{a}})-{\sqrt {a^{2}+{\frac {1}{a^{2}}}}})}}\leq {\frac {2-{\sqrt {2}}}{(2+{\sqrt {2}})(2-{\sqrt {2}})}}\\\Rightarrow {\frac {(a+{\frac {1}{a}})-{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}}{(a+{\frac {1}{a}})^{2}-({\sqrt {a^{2}+{\frac {1}{a^{2}}}}})^{2}}}\leq {\frac {2-{\sqrt {2}}}{2^{2}-({\sqrt {2}})^{2}}}\\\Rightarrow {\frac {a+{\frac {1}{a}}-{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}}{(a^{2}+2+{\frac {1}{a^{2}}})-(a^{2}+{\frac {1}{a^{2}}})}}\leq {\frac {2-{\sqrt {2}}}{4-2}}\\\Rightarrow {\frac {a+{\frac {1}{a}}-{\sqrt {a^{2}+{\frac {1}{a^{2}}}}}}{2}}\leq {\frac {2-{\sqrt {2}}}{2}}\\\Rightarrow {\sqrt {a^{2}+{\frac {1}{a^{2}}}}}-{\sqrt {2}}\geq a+{\frac {1}{a}}-2\end{array}}}$

${\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}{\frac {1}{a}}+{\frac {1}{b}}\geq 2{\sqrt {{\frac {1}{a}}{\frac {1}{b}}}}=2{\frac {1}{\sqrt {ab}}}\\a+b\geq 2{\sqrt {ab}}\end{array}}\right.\\\Rightarrow {\frac {1}{a}}+{\frac {1}{b}}\geq 2{\frac {1}{\sqrt {ab}}}\geq 2\times {\frac {2}{a+b}}={\frac {4}{a+b}}\end{array}}}$

${\displaystyle \left\{{\begin{array}{l}{\frac {1}{a}}+{\frac {1}{b}}\geq {\frac {4}{a+b}}\\{\frac {1}{b}}+{\frac {1}{c}}\geq {\frac {4}{b+c}}\\{\frac {1}{c}}+{\frac {1}{a}}\geq {\frac {4}{c+a}}\end{array}}\right.}$

${\displaystyle {\begin{array}{l}({\frac {1}{a}}+{\frac {1}{b}})+({\frac {1}{b}}+{\frac {1}{c}})+({\frac {1}{c}}+{\frac {1}{a}})\geq {\frac {4}{a+b}}+{\frac {4}{b+c}}+{\frac {4}{c+a}}\\\Rightarrow 2({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}})\geq 4({\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{c+a}})\\\Rightarrow {\frac {1}{2}}({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}})\geq ({\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{c+a}})\end{array}}}$

${\displaystyle {\begin{array}{l}({\frac {1}{2a}}+{\frac {1}{2b}}+{\frac {1}{2c}})^{2}=({\frac {1}{2a}}+{\frac {1}{2b}}+{\frac {1}{2c}})({\frac {1}{2a}}+{\frac {1}{2b}}+{\frac {1}{2c}})\\\geq ({\frac {1}{2{\sqrt {ab}}}}+{\frac {1}{2{\sqrt {bc}}}}+{\frac {1}{2{\sqrt {ca}}}})^{2}\\\geq ({\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{c+a}})^{2}\end{array}}}$

### 平均值不等式的补充

#### 算术-对数-几何平均值不等式

1957年，B·奥斯透（B. Ostle）和H·L·特尔维利尔（H. L. Terwilliger）提出对于不相等的任意2个正数a和b，存在不等式${\displaystyle {\frac {a-b}{\ln a-\ln b}}>{\sqrt {ab}}}$；1966年，B·C·卡尔森（B. C. Carlson）提出对于不相等的任意2个正数a和b，存在不等式${\displaystyle {\frac {a+b}{2}}>{\frac {a-b}{\ln a-\ln b}}}$[1]。将它们合并到一起，并和算术-几何平均值不等式相联系，我们就得到本节的主角：

${\displaystyle a,b>0,a\neq b}$，则有下列算术-对数-几何平均值不等式arithmetic-logarithmic-geometric mean inequality[1]，简称对数均值不等式A-L-G不等式

${\displaystyle {\frac {a+b}{2}}>{\frac {a-b}{\ln a-\ln b}}>{\sqrt {ab}}}$

${\displaystyle {\frac {t+1}{2}}>{\frac {t-1}{\ln t}}>{\sqrt {t}}}$

• 左半边：${\displaystyle \ln x>{\frac {2(x-1)}{x+1}}}$ (取${\displaystyle t=x,x>1}$)
• 右半边：${\displaystyle x-{\frac {1}{x}}>2\ln x}$ (取${\displaystyle t=x^{2},x>1}$)

${\displaystyle {\begin{array}{l}f(x)\geq f(1)=\ln 1-{\frac {2(1-1)}{1+1}}=0-0=0\quad (x\geq 1)\\\Rightarrow \ln x-{\frac {2(x-1)}{x+1}}\geq 0\quad (x\geq 1)\\\Rightarrow \ln x\geq {\frac {2(x-1)}{x+1}}\quad (x\geq 1)\end{array}}}$

${\displaystyle {\begin{array}{l}f(x)\geq f(1)=0\quad (x\geq 1)\\\Rightarrow x-{\frac {1}{x}}-2\ln x\geq 0\quad (x\geq 1)\\\Rightarrow x-{\frac {1}{x}}\geq 2\ln x\quad (x\geq 1)\end{array}}}$

• ${\displaystyle a=x^{2},b=1}$，则有：${\displaystyle \ln {\frac {2(x-1)}{x+1}}<\ln x
• ${\displaystyle a=e^{x_{1}},b=e^{x_{2}}}$，则分别有${\displaystyle {\frac {x_{1}-x_{2}}{2}}>{\frac {e^{x_{1}}-e^{x_{2}}}{e^{x_{1}}+e^{x_{2}}}}}$${\displaystyle x_{1}-x_{2}<{\frac {e^{x_{1}}-x^{x_{2}}}{e^{\frac {x_{1}+x_{2}}{2}}}}}$成立。

A-L-G与中国大陆高考导数压轴题中的极值点偏移考点联系密切。本小节我们只讨论极值点偏移问题的不等式解法（即变形后套用A-L-G不等式解决此类问题），暂不提及其它传统的导数论证方法。

（出自2013年中国大陆高考理科数学陕西卷第21题第(3)问（压轴解答题）。）

${\displaystyle f(x_{1})=f(x_{2})\quad \Rightarrow \quad {\frac {\ln x_{1}}{x_{1}}}={\frac {\ln x_{2}}{x_{2}}}}$

${\displaystyle {\begin{array}{l}{\frac {\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}}={\frac {\ln x_{1}}{x_{1}}}0)\\\Rightarrow {\frac {x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}}>e\end{array}}}$

${\displaystyle {\begin{array}{l}{\frac {x_{1}+x_{2}}{2}}>{\frac {x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}}>e\\\Rightarrow x_{1}+x_{2}>2e\end{array}}}$

${\displaystyle x_{1}+x_{2}>2e\quad \Leftrightarrow \quad {\frac {x_{1}+x_{2}}{2}}>e}$

${\displaystyle {\begin{array}{l}{\sqrt {x_{1}x_{2}}}>e\\\Rightarrow x_{1}x_{2}>e^{2}\\\Rightarrow \ln(x_{1}x_{2})>\ln(e^{2})\\\Rightarrow \ln x_{1}+\ln x_{2}>2\ln e=2\end{array}}}$

${\displaystyle {\frac {\ln x_{1}}{x_{1}}}={\frac {\ln x_{2}}{x_{2}}}}$

${\displaystyle \left\{{\begin{array}{l}\ln t=\ln({\frac {x_{1}}{x_{2}}})=\ln x_{1}-\ln x_{2}\\t={\frac {x_{1}}{x_{2}}}={\frac {\ln x_{1}}{\ln x_{2}}}\end{array}}\right.}$

${\displaystyle \left\{{\begin{array}{l}\ln t=\ln x_{1}-\ln x_{2}\\t={\frac {\ln x_{1}}{\ln x_{2}}}\end{array}}\right.}$

${\displaystyle {\begin{array}{l}\ln x_{1}+\ln x_{2}>2\\\Rightarrow {\frac {t\ln t}{t-1}}+{\frac {\ln t}{t-1}}>2\\\Rightarrow {\frac {(t+1)\ln t}{t-1}}>2\\\Rightarrow \ln t>{\frac {2(t-1)}{t+1}}\quad (0

${\displaystyle x_{1}易知：
${\displaystyle {\frac {\ln x_{1}}{x_{1}}}={\frac {\ln x_{2}}{x_{2}}}

${\displaystyle \left\{{\begin{array}{l}x_{1}=a\ln x_{1}\\x_{2}=a\ln x_{2}\\{\frac {x_{1}}{x_{2}}}=t\end{array}}\right.}$

${\displaystyle \left\{{\begin{array}{l}e^{x_{1}}-2x_{1}-a\\e^{x_{2}}-2x_{2}-a\end{array}}\right.\quad \Rightarrow \quad e^{x_{1}}-e^{x_{2}}=2(x_{1}-x_{2})\quad \Rightarrow \quad 2={\frac {e^{x_{1}}-e^{x_{2}}}{x_{1}-x_{2}}}}$

${\displaystyle 2={\frac {e^{x_{1}}-e^{x_{2}}}{x_{1}-x_{2}}}<{\frac {e^{x_{1}}+e^{x_{2}}}{2}}\quad \Rightarrow \quad 4

${\displaystyle {\begin{array}{l}x_{1}x_{2}

${\displaystyle {\sqrt {(x_{1}-1)(x_{2}-1)}}<{\frac {(x_{1}-1)-(x_{2}-1)}{\ln(x_{1}-1)-\ln(x_{2}-1)}}={\frac {x_{1}-x_{2}}{\ln(x_{1}-1)-\ln(x_{2}-1)}}}$

${\displaystyle \left\{{\begin{array}{l}e^{x_{1}}-ax_{1}+a=0\\e^{x_{2}}-ax_{2}+a=0\end{array}}\right.\quad \Rightarrow \quad \left\{{\begin{array}{l}e^{x_{1}}=a(x_{1}-1)\\e^{x_{2}}=a(x_{2}-1)\end{array}}\right.\quad \Rightarrow \quad \left\{{\begin{array}{l}{\frac {e^{x_{1}}}{a}}=x_{1}-1\\{\frac {e^{x_{2}}}{a}}=x_{2}-1\end{array}}\right.}$
${\displaystyle {\begin{array}{l}{\sqrt {(x_{1}-1)(x_{2}-1)}}<{\frac {(x_{1}-1)-(x_{2}-1)}{\ln(x_{1}-1)-\ln(x_{2}-1)}}={\frac {x_{1}-x_{2}}{\ln({\frac {e^{x_{1}}}{a}})-\ln({\frac {e^{x_{2}}}{a}})}}\\={\frac {x_{1}-x_{2}}{(\ln e^{x_{1}}-\ln a)-(\ln e^{x_{2}}-\ln a)}}={\frac {x_{1}-x_{2}}{x_{1}-\ln a-x_{2}+\ln a}}={\frac {x_{1}-x_{2}}{x_{1}-x_{2}}}=1\end{array}}}$

### 排序不等式

${\displaystyle a_{1}\leqslant a_{2}\leqslant \cdots \leqslant a_{n},b_{1}\leqslant b_{2}\leqslant \cdots \leqslant b_{n},\{c_{1},c_{2},\cdots ,c_{n}\}=\{b_{1},b_{2},\cdots ,b_{n}\}}$，那么有

${\displaystyle \sum _{i=1}^{n}a_{i}b_{i}\geqslant \sum _{i=1}^{n}a_{i}c_{i}\geqslant \sum _{i=1}^{n}a_{i}b_{n+1-i}}$

#### 切比雪夫不等式

${\displaystyle a_{1}\leqslant a_{2}\leqslant \cdots \leqslant a_{n},b_{1}\leqslant b_{2}\leqslant \cdots \leqslant b_{n}}$，那么

${\displaystyle \sum _{i=1}^{n}a_{i}b_{n+1-i}\leqslant {\frac {1}{n}}\left(\sum _{i=1}^{n}a_{i}\right)\left(\sum _{i=1}^{n}b_{i}\right)\leqslant \sum _{i=1}^{n}a_{i}b_{i}}$

{\displaystyle {\begin{aligned}&\sum _{i=1}^{n}a_{i}b_{i}\geqslant \sum _{i=1}^{n}a_{i}b_{i}\\&\sum _{i=1}^{n}a_{i}b_{i}\geqslant \sum _{i=1}^{n}a_{i}b_{i+1}\\&\sum _{i=1}^{n}a_{i}b_{i}\geqslant \sum _{i=1}^{n}a_{i}b_{i+2}\\&\vdots \\&\sum _{i=1}^{n}a_{i}b_{i}\geqslant \sum _{i=1}^{n}a_{i}b_{i+n-1}\end{aligned}}}

${\displaystyle n\sum _{i=1}^{n}a_{i}b_{i}\geqslant \left(\sum _{i=1}^{n}a_{i}\right)\left(\sum _{i=1}^{n}b_{i}\right)}$

### 上凸与下凸函数

${\displaystyle f\left({\frac {x_{1}+x_{2}}{2}}\right)\leqslant {\frac {1}{2}}[f(x_{1})+f(x_{2})]}$

### 延森不等式

#### 延森不等式及其证明

${\displaystyle \forall x_{1},x_{2},x_{3},...,x_{n}\in I,{\frac {\sum _{i=1}^{n}f(x_{i})}{n}}\geq f({\frac {\sum _{i=1}^{n}x_{i}}{n}})}$

${\displaystyle \forall x_{1},x_{2},x_{3},...,x_{n}\in I,\sum _{i=1}^{n}\lambda _{i}=1,\sum _{i=1}^{n}\lambda _{i}f(x_{i})\geq f(\sum _{i=1}^{n}\lambda _{i}x_{i})}$

#### 延森不等式的简单应用

• ${\displaystyle f(x)=\sin x\quad (x\in [0,\pi ])}$
• ${\displaystyle f(x)=\cos x\quad (x\in [0,{\frac {\pi }{2}}]\ or\ [{\frac {\pi }{2}},\pi ])}$
• ${\displaystyle f(x)=\tan x\quad (x\in [0,{\frac {\pi }{2}}))}$
• ${\displaystyle f(x)=\ln x}$${\displaystyle f(x)=lgx\quad (x>0)}$
• ${\displaystyle f(x)=x\ln x}$${\displaystyle f(x)=xlgx\quad (x>0)}$

${\displaystyle {\frac {f(a)+f(b)}{2}}\geq f({\frac {a+b}{2}})=f({\frac {1}{2}})}$，当且仅当${\displaystyle a=b={\frac {1}{2}}}$时等号成立。

${\displaystyle {\frac {{\sqrt {1+a^{2}}}+{\sqrt {1+b^{2}}}}{2}}\geq {\sqrt {1+({\frac {1}{2}})^{2}}}={\frac {\sqrt {5}}{2}}\Leftrightarrow {\sqrt {1+a^{2}}}+{\sqrt {1+b^{2}}}\geq {\sqrt {5}}}$

${\displaystyle f(x)={\sqrt {x+5}}\quad (x\geq -5)}$，易得${\displaystyle f'(x)=-{\frac {1}{\sqrt {x+5}}}}$

${\displaystyle {\begin{array}{l}{\frac {f(a)+f(b)+f(c)}{3}}\leq f({\frac {a+b+c}{3}})\\\Rightarrow {\frac {{\sqrt {a+5}}+{\sqrt {b+5}}+{\sqrt {c+5}}}{3}}\leq {\sqrt {{\frac {a+b+c}{3}}+5}}\end{array}}}$

${\displaystyle {\begin{array}{l}{\sqrt {a+5}}+{\sqrt {b+5}}+{\sqrt {c+5}}\leq 3{\sqrt {{\frac {a+b+c}{3}}+5}}\\=3{\sqrt {{\frac {1}{3}}+5}}=3{\sqrt {\frac {16}{3}}}=4{\sqrt {3}}\end{array}}}$

${\displaystyle \left\{{\begin{array}{l}k_{1}\cdot {\sqrt {a+5}}\leq {\frac {k_{1}^{2}+(a+5)}{2}}\quad (k_{1}>0)\\k_{2}\cdot {\sqrt {b+5}}\leq {\frac {k_{2}^{2}+(b+5)}{2}}\quad (k_{2}>0)\\k_{3}\cdot {\sqrt {c+5}}\leq {\frac {k_{3}^{2}+(c+5)}{2}}\quad (k_{3}>0)\end{array}}\right.}$

${\displaystyle {\begin{array}{l}k_{1}{\sqrt {a+5}}+k_{2}{\sqrt {b+5}}+k_{3}{\sqrt {c+5}}\\\leq {\frac {k_{1}^{2}+(a+5)}{2}}+{\frac {k_{2}^{2}+(b+5)}{2}}+{\frac {k_{3}^{2}+(c+5)}{2}}\geq 4{\sqrt {3}}\\={\frac {k_{1}^{2}+k_{2}^{2}+k_{3}^{2}}{2}}+{\frac {(a+b+c)+15}{2}}\geq 4{\sqrt {3}}\end{array}}}$

${\displaystyle {\begin{array}{l}{\sqrt {\frac {16}{3}}}\cdot {\sqrt {a+5}}+{\sqrt {\frac {16}{3}}}\cdot {\sqrt {b+5}}+{\sqrt {\frac {16}{3}}}\cdot {\sqrt {c+5}}\leq {\frac {({\sqrt {\frac {16}{3}}})^{2}+({\sqrt {\frac {16}{3}}})^{2}+({\sqrt {\frac {16}{3}}})^{2}}{2}}+{\frac {(a+b+c)+15}{2}}\\\Rightarrow {\frac {\sqrt {16}}{3}}({\sqrt {a+5}}+{\sqrt {b+5}}+{\sqrt {c+5}})\leq {\frac {3\cdot {\frac {16}{3}}}{2}}+{\frac {(a+b+c)+15}{2}}\\\Rightarrow {\sqrt {a+5}}+{\sqrt {b+5}}+{\sqrt {c+5}}\leq (8+{\frac {(a+b+c)+15}{2}})\cdot {\sqrt {\frac {3}{16}}}\end{array}}}$

${\displaystyle {\begin{array}{l}{\sqrt {a+5}}+{\sqrt {b+5}}+{\sqrt {c+5}}\leq (8+{\frac {1+15}{2}})\cdot {\sqrt {\frac {3}{16}}}\\=16\cdot {\sqrt {\frac {3}{16}}}={\sqrt {3\cdot 16}}=4{\sqrt {3}}\end{array}}}$

#### 需要进行对数变换的问题

• 定义在实数向量空间中凸集内，函数值为正数。
• 函数进行对数变换后为（下）凸函数。

${\displaystyle \ln({\frac {1}{x^{2}}}+x)+\ln({\frac {1}{y^{2}}}+y)+\ln({\frac {1}{z^{2}}}+z)\geq 3\ln({\frac {28}{3}})}$

${\displaystyle {\frac {f(x)+f(y)+f(z)}{3}}\geq f({\frac {x+y+z}{3}})=f({\frac {1}{3}})}$，当且仅当${\displaystyle x=y=z={\frac {1}{3}}}$时等号成立。

${\displaystyle {\frac {\ln({\frac {1}{x^{2}}}+x)+\ln({\frac {1}{y^{2}}}+y)+\ln({\frac {1}{z^{2}}}+z)}{3}}\geq \ln({\frac {1}{({\frac {1}{3}})^{2}}}+({\frac {1}{3}}))}$
${\displaystyle \ln({\frac {1}{x^{2}}}+x)+\ln({\frac {1}{y^{2}}}+y)+\ln({\frac {1}{z^{2}}}+z)\geq 3\ln({\frac {1}{({\frac {1}{3}})^{2}}}+({\frac {1}{3}}))=3\ln({\frac {1}{\frac {1}{9}}}+{\frac {1}{3}})=\ln(9+{\frac {1}{3}})^{3}=\ln({\frac {28}{3}})^{3}}$
${\displaystyle \ln(({\frac {1}{x^{2}}}+x)({\frac {1}{y^{2}}}+y)({\frac {1}{z^{2}}}+z))\geq \ln({\frac {28}{3}})^{3}}$，即${\displaystyle ({\frac {1}{x^{2}}}+x)({\frac {1}{y^{2}}}+y)({\frac {1}{z^{2}}}+z)\geq \ln({\frac {28}{3}})^{3}}$。证明完毕。

${\displaystyle {\frac {x+y}{a+b}}\ln({\frac {a}{a+b}}{\frac {x}{a}}+{\frac {b}{a+b}}{\frac {y}{b}})\leq {\frac {a}{a+b}}({\frac {x}{a}})\ln({\frac {x}{a}})+{\frac {b}{a+b}}({\frac {y}{b}})\ln({\frac {y}{b}})}$

${\displaystyle f(\lambda _{1}({\frac {x}{a}})+\lambda _{2}({\frac {y}{b}}))\leq \lambda _{1}f({\frac {x}{a}})+\lambda _{2}f({\frac {y}{b}}))}$，当且仅当${\displaystyle {\frac {x}{a}}={\frac {y}{b}}}$时等号成立。

${\displaystyle {\begin{array}{l}(\lambda _{1}({\frac {x}{a}})+\lambda _{2}({\frac {y}{b}}))\ln(\lambda _{1}({\frac {x}{a}})+\lambda _{2}({\frac {y}{b}}))\leq \lambda _{1}({\frac {x}{a}})\ln({\frac {x}{a}})+\lambda _{2}({\frac {y}{b}})\ln({\frac {y}{b}})\\\Leftrightarrow (({\frac {a}{a+b}})({\frac {x}{a}})+({\frac {b}{a+b}})({\frac {y}{b}}))\ln(({\frac {a}{a+b}})({\frac {x}{a}})+({\frac {a}{a+b}})({\frac {y}{b}}))\leq ({\frac {a}{a+b}})({\frac {x}{a}})\ln({\frac {x}{a}})+({\frac {b}{a+b}})({\frac {y}{b}})\ln({\frac {y}{b}})\\\Leftrightarrow (({\frac {x}{a+b}})+({\frac {y}{a+b}})\ln({\frac {x}{a+b}}+{\frac {y}{a+b}})\leq {\frac {x}{a+b}}\ln {\frac {x}{a}}+{\frac {y}{a+b}}\ln {\frac {y}{b}}\\\Leftrightarrow (({\frac {x}{a+b}})+({\frac {y}{a+b}})\ln {\frac {x+y}{a+b}}\leq {\frac {x}{a+b}}\ln {\frac {x}{a}}+{\frac {y}{a+b}}\ln {\frac {y}{b}}\\\Leftrightarrow ((x+y)\ln {\frac {x+y}{a+b}}\leq x\ln {\frac {x}{a}}+y\ln {\frac {y}{b}}\end{array}}}$

${\displaystyle (am+bn)\ln {\frac {a^{2}m+b^{2}n}{a+b}}\leq am\ln m+bn\ln n}$

${\displaystyle {\frac {am+bn}{a+b}}\ln {\frac {a^{2}m+b^{2}n}{a+b}}\leq {\frac {a}{a+b}}(m\ln m)+{\frac {b}{a+b}}(n\ln n)}$

${\displaystyle f(\lambda _{1}m+\lambda _{2}n)\leq \lambda _{1}f(m)+\lambda _{2}f(n)}$，当且仅当${\displaystyle m=n}$（即${\displaystyle {\frac {x}{a}}={\frac {y}{b}}}$）时等号成立。

${\displaystyle {\begin{array}{l}(\lambda _{1}m+\lambda _{2}n)\ln(\lambda _{1}m+\lambda _{2}n)\leq \lambda _{1}m\ln m+\lambda _{2}n\ln n\\\Leftrightarrow ({\frac {a}{a+b}}m+{\frac {b}{a+b}}n)\ln({\frac {a}{a+b}}m+{\frac {b}{a+b}}n)\leq {\frac {a}{a+b}}m\ln m+{\frac {b}{a+b}}n\ln n\\\Leftrightarrow ({\frac {am}{a+b}}+{\frac {bn}{a+b}})\ln {\frac {am+bn}{a+b}}\leq {\frac {a}{a+b}}m\ln m+{\frac {b}{a+b}}n\ln n\end{array}}}$

## 参考资料

1. （英文）József Sándor（2015年）．A Basic Logarithmic Inequality, And The Logarithmic Mean（pdf）．Notes on Number Theory and Discrete Mathematics，31-35．