Intro

对数均值不等式（Arithmetic-Logarithmic-Geometric mean inequalities，ALG不等式）

当$a>0, b>0$时有：

$\color{grey}{b>\sqrt{\dfrac{a^{2}+b^{2}}{2}}>}$$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}>\sqrt{a b}$$\color{grey}{>\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}>a}$

其中$\dfrac{b-a}{\ln b-\ln a}$为对数均值。

证明

核心思路：令$t=\dfrac{a}{b}$，构造关于$t$的方程$f(t).$

证明右半边 : 当$a > 0 , b > 0$, 且$a ≠ b$时 , 有$\sqrt{ab}< \dfrac{a-b}{ \ln a- \ln b}$.
证明：不妨设$a > b$, 上不等式⇔$\ln \dfrac{a}{b}< \dfrac{a-b}{ \sqrt{ab}}= \sqrt{ \dfrac{a}{b}}- \sqrt{ \dfrac{b}{a}}$.
令$t= \sqrt{ \dfrac{a}{b}}\in (1,+ \infty)$，
上不等式⇔ $2 \ln t < t - \dfrac{1}{t}$ $\Leftrightarrow t^{2}+2t \ln t>1$.
构造函数$f ( t ) = t ^2+2 t\ln t$, 求导得 : $f ' ( t ) = 2 t +2\ln t +2$.
当$t ∈ ( 1 , + ∞ )$时 , $f ′ ( t ) > 0$恒成立, 即可知$f(t)$在$( 1 , + ∞)$上单调递增 .
即有$f ( t ) > f ( 1 ) = 1$, 所以原不等式成立 .
左半边同理。

套路

证明ALG不等式；
根据$f(x_1)=f(x_2)=a$联立2个等式；
指数化对数，有$a$的相减消$a$；
得到关系，代入具体题目分析。
*如有需要，方程组可以相加，以得到$a$与$x_1,x_2$的关系。

推论

全部导数：已知$0 < b < a$，
则: $\dfrac{1}{b}>\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{2}>\dfrac{1}{\sqrt{a b}}>\dfrac{\ln a-\ln b}{a-b}>\dfrac{2}{a+b}>\sqrt{\dfrac{2}{a^{2}+b^{2}}}>\dfrac{1}{a}$
推论1：
$\begin{array}{ll}x>1 \text { 时, } & 2 \dfrac{x-1}{x+1}<\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right) \\ x \in(0,1) \text { 时, } & 2 \dfrac{x-1}{x+1}>\ln x>\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\end{array}$
推论2：
$\begin{array}{ll}x>1 \text { 时, } & \dfrac{x^{2}-1}{x^{2}+1}<\ln x<\sqrt{x}-\dfrac{1}{\sqrt{x}} \\ x \in(0,1) \text { 时 }, & \dfrac{x^{2}-1}{x^{2}+1}>\ln x>\sqrt{x}-\dfrac{1}{\sqrt{x}}\end{array}$

实战演练

小试牛刀

已知 $f(x)=\ln x-a x$有两个零点$x_{1}, x_{2}$.
1. 求a的取值范围; (答案：$a \in\left(0,\frac{1}{e}\right)$)
2. 求证 $x_{1} \cdot x_{2}>e^{2}$.
解析
设2个零点为$x_{1}, x_{2}$. 则有
$\left\{\begin{array}{l}\ln x_{1}=a x_{1} \\ \ln x_{2}=a x_{2}\end{array}\right.$
$\ln x_{1}-\ln x_{2}=a\left(x_{1}-x_{2}\right)$
$\dfrac{1}{a}=\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}<\dfrac{x_{1}+x_{2}}{2}$
$\therefore a >x_{1} \dfrac{2}{x_{1}+x_{2}}$
$\ln x_{1}+\ln x_{2}=a\left(x_{1}+x_{2}\right)$ $>\dfrac{2}{x_{1}+x_{2}} \left(x_{1}+x_{2}\right)=2$
$\therefore\ln \left(x_{1} x_{2}\right)>2$,$x_{1} x_{2}>e^{2}$
证明: $x-\ln x>2 \dfrac{e^{x}-x}{e^{x}+x}$.
解析
$\dfrac{\ln e^{x}-\ln x}{e^{x}-x}>\dfrac{2}{e^{x}+x}$
$x-\ln x>2 \dfrac{e^{x}-x}{e^{x}+x}$
已知 $f(x)=\ln x-a\left(x-\dfrac{1}{x}\right)$,对于$\forall x \in(1,+\infty)$都有$f(x)<0$恒成立，则正数$a$的取值范围是($\qquad$).
A.$(0,2]\qquad$ B.$[2,+\infty)\qquad$ C. $\left(0, \dfrac{1}{2}\right]\qquad$ D.$\left[\dfrac{1}{2},+\infty\right)$
解析
D. 解析：$x>1 , \ln x<\dfrac{1}{2}\left(x-\dfrac{1}{a}\right) \leq a\left(x-\dfrac{1}{x}\right)$ $\Rightarrow a \geqslant \dfrac{1}{2}$
已知 $f(x)=\ln (x+1)-\dfrac{a x}{x+a}$,对于$\forall x \in(0,+\infty)$都有$f(x)>0$恒成立，则正数a的取值范围是($\qquad$).
A.$(0,2]\qquad$ B.$[2,+\infty)\qquad$ C.$\left(0, \dfrac{1}{2}\right]\qquad$D.$\left[\dfrac{1}{2},+\infty\right)$
解析
A. 解析：$\ln x>2 \dfrac{x-1}{x+1} \quad(x>1)$
把 $x=x+1$ 代入得：
$x>0$时, $\ln (x+1)>2 \dfrac{x+1-1}{x+1+1}=\dfrac{2x}{x+2}$
$\ln (x+1)>\dfrac{2 x}{x+2} \geqslant \dfrac{ax}{x+a}$
$2 x+2 a \geqslant a x+2 a$
$\therefore a \leqslant 2$

大显身手

(2016全国课标Ⅰ卷理科21题)已知函数$f(x)=(x-2)e^{x}+a(x-1)$有两个零点.
(Ⅰ) 求$a$的取值范围 ;
(Ⅱ) 设$x _1 , x _2$是$f ( x )$的两个零点 , 证明: $x _1+ x _2< 2$.
解析
解析 : (Ⅰ) $a∈(0，+∞)$
(Ⅱ) 由 (Ⅰ) 知$a > 0 , x _1 < 1 <x _2 < 2 , f ( x _1 ) = f ( x _2 ) = 0 ,$
所以$\begin{cases} a(x_{1}-1)^{2}=(2-x_{1})e^{x} \\ a(x_{2}-1)^{2}=(2-x_{2})e^{x_{2}} \end{cases}$
取对数可得$\begin{cases} \ln a+2 \ln (1-x_{1})= \ln (2-x_{1})+x_{1} \\ \ln a+2 \ln (x_{2}-1)= \ln (2-x_{2})+x_{2} \end{cases}$
相减可得$2 ( 1- x _1 ) -2 ( x _2-1 ) = ( 2- x _1 ) -( 2- x _2 ) ＋ x _1- x _2 $,
即$2 \ln ( 1- x _1 ) -2 \ln ( x _2-1 ) = \ln ( 2- x _1 )- \ln (2-x_{2})- \left[ (2-x_{1})-(2-x_{2}) \right]$,
所以$\dfrac{2 \ln (1-x_{1})-2 \ln (x_{2}-1)}{ \ln (2-x_{1})- \ln (2-x_{2})}= 1- \dfrac{(2-x_{1})-(2-x_{2})}{ \ln (2-x_{1})- \ln (2-x_{2})}$,
若$x _1 ＋ x _2 \geqslant 2$, 则$1-x _1 \leqslant x _2-1$,
所以$\dfrac{2 \ln (1-x_{1})-2 \ln (x_{2}-1)}{ \ln (2-x_{1})- \ln (2-x_{2})} \leqslant 0$,
所以$1 \leqslant \dfrac{(2-x_{1})-(2-x_{2})}{ \ln (2-x_{1})- \ln (2-x_{2})}<\dfrac{(2-x_{1})+(2-x_{2})}{2}$,
所以$( 2- x _1 ) + ( 2 - x _2 ) > 2$, 即$x _1+ x _2 < 2$, 这与$x _1+ x _2 ≥ 2$矛盾 , 所以$x _1+ x _2 < 2$.

已知函数 $f(x)=2 \ln x-x^{2}-a x$.
1. 求函数 $f(x)$ 的单调区间;
2. 如果函数 $f(x)$ 有两个不同的零点 $x_{1}, x_{2},$且$x_{1}<x_{2},$证明:$f'\left(\dfrac{x_{1}+x_{2}}{2}\right)<0$
解析
分析： (1) 单调递增区间为 $\left(0, \dfrac{-a+\sqrt{a^{2}+16}}{4}\right)$, 单调递减区间为 $\left(\dfrac{-a+\sqrt{a^{2}+16}}{4},+\infty\right)$
1. 证： $\because f(x)=2 \ln x-x^{2}-a x(x>0) \therefore f^{\prime}(x)=\dfrac{2}{x}-2 x-a$,
由题意得：$\left\{\begin{array}{l}2 \ln x_{1}-x_{1}^{2}-a x_{1}=0 \\ 2 \ln x_{2}-x_{2}^{2}-a x_{2}=0\end{array}\right.$,
两式相减得 $: 2\left(\ln x_{1}-\ln x_{2} \right) -\left(x_{1}^{2}-x_{2}^{2}\right)=a\left(x_{1}-x_{2}\right)$
$\Leftrightarrow\dfrac{2\left(\ln x_{1}-\ln x_{2}\right)-\left(x_{1}^{2}-x_{2}^{2}\right)}{x_{1}-x_{2}}=a$,
欲证 $: f^{\prime}\left(\dfrac{x_{1}+x_{2}}{2}\right)<0$
只需证 $: f^{\prime}\left(\dfrac{x_{1}+x_{2}}{2}\right)=\dfrac{4}{x_{1}+x_{2}}-\left(x_{1}+x_{2}\right)-a<0$,
只需证： $\dfrac{4}{x_{1}+x_{2}}-\left(x_{1}+x_2\right)<a$,
只需证： $\dfrac{4}{x_{1}+x_{2}}-\left(x_{1}+x_2\right)<\dfrac{2\left(\ln x_{1}-\ln x_{2}\right)-\left(x_{1}^{2}-x_{2}^{2}\right)}{x_{1}-x_{2}}$,
只需证：$\dfrac{2}{x_{1}+x_{2}}<\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}$,
由对数均值不等式$\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}<\dfrac{x_{1}+x_{2}}{2}$可得$\dfrac{2}{x_{1}+x_{2}}<\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}$,
从而命题得证.
(2014南通二模)设函数 $f(x)=e^{x}-a x+a$ $(a \in \mathbf{R})$, 其图象与$x$轴交于$A\left(x_{1}, 0\right)$,$B\left(x_{2}, 0\right)$ 两点, 且 $x_{1}<x_{2}$.
1. 求实数a 的取值范围;
2. 证明：$f'(\sqrt{x_{1} x_{2}})<0 .$
解析
解：(1) $a>e^{2}$ (过程略)
证：(2) $\because f(1)=e>0,$由$(1)$知$1<x_{1}<\ln a<x_{2}$
由题意得： $\left\{\begin{array}{l}e^{x_1} -a x_{1}+a=0\\ e^{x_2}-a x_{2}+a=0\end{array} \Leftrightarrow\left\{\begin{array}{l}e^{x_{1}}=a\left(x_{1}-1\right) \\ e^{x_{2}}=a\left(x_{2}-1\right)\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=\ln \left(x_{1}-1\right)+\ln a \\ x_{2}=\ln \left(x_{2}-1\right)+\ln a\end{array}\right.\right.\right.$
两式相减得 $x_{1}-x_{2}=\ln (x-1)-\ln \left(x_{2}-1\right),$
$\therefore \dfrac{x_{1}-x_{2}}{\ln \left(x_{1}-1\right)-\ln \left(x_{2}-1\right)}=$ $\dfrac{\left(x_{1}-1\right)-\left(x_{2}-1\right)}{\ln \left(x_{1}-1\right)-\ln \left(x_{2}-1\right)}=1$
由ALG不等式得： $\sqrt{\left(x_{1}-1\right)\left(x_{2}-1\right)}<\dfrac{\left(x_{1}-1\right)-\left(x_{2}-1\right)}{\ln \left(x_{1}-1\right)-\ln \left(x_{2}-1\right)}$
$\therefore \sqrt{\left(x_{1}-1\right)}\left(x_{2}-1\right)<1, \therefore\left(x_{1}-1\right)\left(x_{2}-1\right)<1$

由 $\left\{\begin{array}{l}x_{1}=\ln \left(x_{1}-1\right)+\ln a \\ x_{2}=\ln \left(x_{2}-1\right)+\ln a\end{array}\right.$
两式相加得：$x_{1}+x_{2}=\left[\ln \left(x_{1}-1\right)+\ln \left(x_{2}-1\right)\right]+$ $2 \ln a$
即 $\left(x_{1}+x_{2}\right)-2 \ln a=\ln \left[\left(x_{1}-1\right) \left(x_{2}-1\right)\right],$
$\because\left(x_{1}-1\right) \left(x_{2}-1\right)<1 \therefore \ln \left[\left(x_{1}-1\right)\right.\left.\left(x_{2}-1\right)\right]<0$,
$\therefore\left(x_{1}+x_{2}\right)-2 \ln a<0 \therefore \dfrac{x_{1}+x_{2}}{2}<\ln a$,
又$\because \sqrt{x_{1} x_{2}}<\dfrac{x_{1}+x_{2}}{2} \therefore \sqrt{x_{1} x_{2}}<\ln a$ 由(1)知$f(x)$在 $(-\infty, \ln \alpha)$ 上单调递减,
$\therefore f^{\prime}\left(\sqrt{x_{1} x_{2}}\right)<f^{\prime}(\ln \alpha)=0$,
则 $f^{\prime}\left(\sqrt{x_{1} x_{2}}\right)<0$ 得证.
（2014陕西理21(3)）设函数$f(x)=\ln (1+x), g(x)=x f^{\prime}(x)(x \geqslant 0),$其中$f'(x)$是$f(x)$的导函数. 设 $n \in N_{+}$, 比较$g(1)+g(2)+\cdots+g(n)$与 $n-f(n)$ 的大小, 并加以证明.
解析
因为 $g(x)=\dfrac{x}{1+x}$,
所以 $g(1)+g(2)+$ $\cdots+g(n)$ $=\dfrac{1}{2}+\dfrac{2}{3}+\cdots+\dfrac{n}{n+1}$ $=n-\left(\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n+1}\right)$,
而 $n-f(n)=n-\ln (n+1)$, 因此只需比较$\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n+1}$ 与 $\ln (n+1)$ 的大小即可.
利用对数均值不等式: 当 $b>a>0$ 时,有$b>\dfrac{b-a}{\ln b-\ln a} \Rightarrow \ln b-\ln a>\dfrac{b-a}{b}$^{$\color{grey}[注1]$},
故可命$b=n+1, a=$$n,$
则 $\ln (n+1)-\ln n>\dfrac{1}{n+1}$,
故 $\ln 2-\ln 1>\dfrac{1}{2}$,$\ln 3-\ln 2>\dfrac{1}{3}, \cdots, \ln (n+1)-\ln n>\dfrac{1}{n+1},$
将以上各不等式左右两边分别相加得
$\ln (n+1)>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}$$+\cdots+\dfrac{1}{n}+\dfrac{1}{n+1},$
即得 $g(1)+g(2)+\cdots+g(n)>$ $n-f(n)$
注1
【注1】这里如果用$b>\dfrac{b-a}{\ln b-\ln a}$可以很快推出$\ln (n+1)-\ln n>\dfrac{1}{n+1}$. 但是如果使用常规（常用）的ALG不等式也可以分步推出相同结论。
令$b=n+1, a=$$n,$根据ALG不等式，有：
$\dfrac{(n+1)-n}{\ln(n+1)-\ln n}<\dfrac{n+1+n}{2}$ $\dfrac{1}{\ln (n+1)-\ln n}<\dfrac{2 n+1}{2}$ $\ln (n+1)-\ln n>\dfrac{2}{2 n+1}>\dfrac{2}{2 n+2}=\dfrac{1}{n+1}$

（2013 年全国大纲卷22II）设函数 $f(x)=\ln (1+x)-\dfrac{x(1+\lambda x)}{1+x}$. 设数列 $\left\{a_{n}\right\}$ 的通项 $a_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots$ $+\dfrac{1}{n}$, 证明 $: a_{2 n}-a_{n}+\dfrac{1}{4 n}>\ln 2$.
解析
证明: 当 $b>a>0$ 时,有 $\dfrac{b-a}{\ln b-\ln a}>\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}$ (利用对数均值不等式),
即 $\ln b$ $-\ln a<\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(b-a)$,
令 $a=n, b=n+1$, 则 $\ln (n+1)-\ln n<\dfrac{1}{2}\left(\dfrac{1}{n}+\dfrac{1}{n+1}\right),$
所以 $\ln (n+1)$ $-\ln n<\dfrac{1}{2}\left(\dfrac{1}{n}+\dfrac{1}{n+1}\right)$
$\ln (n+2)-\ln (n+1)$$<\dfrac{1}{2}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}\right)$
$\ln (n+3)-\ln (n+2)<$ $\dfrac{1}{2}\left(\dfrac{1}{n+2}+\dfrac{1}{n+3}\right), \cdots,$
$\ln (2 n)-\ln (2 n-1)<$ $\dfrac{1}{2}\left(\dfrac{1}{2 n-1}+\dfrac{1}{2 n}\right)$,
将上式相加可得
$\ln 2 n-\ln n<$ $\dfrac{1}{2}\left(\dfrac{1}{n}+\dfrac{2}{n+1}+\dfrac{2}{n+2}+\dfrac{2}{n+3}+\cdots+\dfrac{2}{2 n-1}+\dfrac{1}{2 n}\right)$ $\ln 2<\dfrac{1}{2 n}+\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+\cdots+\dfrac{1}{2 n-1}\right)$ $+\dfrac{1}{4 n},$
故 $: \dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+\cdots+\dfrac{1}{2 n-1}+\dfrac{1}{2 n}+$ $\dfrac{1}{4 n}>\ln 2,$
所以得证 $a_{2 n}-a_{n}+\dfrac{1}{4 n}>\ln 2$
(2010天津理科21题) 已知函数$f(x)=x e^{-x}(x \in R)$. 如果$x_{1} \neq x_{2}$, 且$f\left(x_{1}\right)=f\left(x_{2}\right)$, 证明$x_{1}+$ $x_{2}>2$.
解析
由 $f\left(x_{1}\right)=f\left(x_{2}\right),$ 得$x_{1} e^{-x_{1}}=x_{2} e^{-x_{2}},$
化简得 $e^{x_{2}-x_{1}}=\dfrac{x_{2}}{x_{1}}$,
两边同时取以 $e$ 为底的对数, 得
$x_{2}-x_{1}$ $=\ln \dfrac{x_{2}}{x_{1}}=\ln x_{2}-\ln x_{1},$也即$\dfrac{\ln x_{2}-\ln x_{1}}{x_{2}-x_{1}}=1,$
利用ALG不等式得 $1=\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}<\dfrac{x_{1}+x_{2}}{2},$
即证得$x_{1}+x_{2}>2$.
(2018全国I卷21) 已知函数 $f(x)=\dfrac{1}{x}-x+$ $a \ln x$. 若$f(x)$存在两个极值点 $x_{1}, x_{2}$, 证明: $\dfrac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}<a-2$.
解析
对函数 $f(x)$ 求导得: $f^{\prime}(x)=\dfrac{a}{x}-\dfrac{1}{x^{2}}-1$ $=\dfrac{-x^{2}+a x-1}{x^{2}}$.
$f(x)$存在两个极值点 $x_{1}, x_{2} \Leftrightarrow$ $-x^{2}+a x-1=0$在$(0,+\infty)$上有两个解$x_{1}, x_{2}$.
由韦达定理可得: $x_{1} \cdot x_{2}=1, x_{1}+x_{2}=a$.
$\left\{\begin{array}{l}f\left(x_{1}\right)=\dfrac{1}{x_{1}}-x_{1}+a \ln x_{1} \\ f\left(x_{2}\right)=\dfrac{1}{x_{2}}-x_{2}+a \ln x_{2}\end{array}\right.$
两式相减得: $f(x_1)-f\left(x_{2}\right)=\dfrac{x_{2}-x_{1}}{x_{2} x_{1}}+x_{2}-x_{1}+a\left(\ln x_{1}-\ln x_{2}\right) .$
结合韦达定理得: $\dfrac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}=-\dfrac{1}{x_{2} x_{1}}-1+$ $a \dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}$ $=a \dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}-2,$
利用ALG不等式可得:$\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}<\dfrac{1}{\sqrt{x_{1} x_{2}}}=1,$
所以 $\dfrac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}<a-2$ 成立.
(2016湖南高联预15) 已知函数 $f(x)=x \ln x-\dfrac{1}{2} m x^{2}-x, m \in \mathbb{R}$. 若 $f(x)$ 有两个极值点 $x_{1}, x_{2}$, 且 $x_{1}<x_{2},$ 求证: $x_{1} x_{2}>e^{2}$.
解析
对 $f(x)$ 求导得: $f^{\prime}(x)=\ln x-mx$.
因为 $x_{1}, x_{2}$为$f(x)$的两个极值点,
所以 $\left\{\begin{array}{l}f^{\prime}\left(x_{1}\right)=\ln x_{1}-m x_{1}=0 \\ f^{\prime}\left(x_{2}\right)=\ln x_{2}-m x_{2}=0\end{array} \Rightarrow\left\{\begin{array}{l}\ln x_{1}=m x_{1} \quad (1)\\ \ln x_{2}=m x_{2}\quad (2)\end{array}\right.\right.$
由(1)-(2)得: $\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}=\dfrac{1}{m}$.
根据对数均值不等式得： \[\dfrac{1}{m}=\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}<\dfrac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}>\dfrac{2}{m}.\quad(3)\]
由(1)+(2)得:
\[\ln \left(x_{1} \cdot x_{2}\right)=m\left(x_{1}+x_{2}\right)\quad(4)\]
将(3)式带入(4)式得: $\ln \left(x_{1} \cdot x_{2}\right)>m \cdot \dfrac{2}{m}=2$ $\Rightarrow x_{1} x_{2}>e^{2}$成立。
(2018福建高联预14) 已知$f(x)=e^{x}-m x .$若$x_{1}, x_{2}$是函数$f(x)$ 的两个零点, 求证:$x_{1}+x_{2}>2$.
解析
因为 $x_{1}, x_{2}$是函数$f(x)$ 的两个零点，
所以$\left\{\begin{array}{l}f\left(x_{1}\right)=e^{x_{1}}-m x_{1}=0 \\ f\left(x_{2}\right)=e^{x_{2}}-m x_{2}=0\end{array}\right.$成立.
移项可得：$\left\{\begin{array}{l}e^{x_{1}}=m x_{1} \\ e^{x_{2}}=m x_{2}\end{array}\right.$
两边分别取对数可得: $\left\{\begin{array}{l}x_{1}=\ln m+\ln x_{1} \\ x_{2}=\ln m+\ln x_{2}\end{array} \quad\right.$
上式相减可得: $\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}=1$ 利用ALG不等式 $\dfrac{2}{x_{1}+x_{2}}<\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}=1$即可得:$x_{1}+x_{2}>2$ 成立.
已知实数 $a>0,$ 函数 $f(x)=\ln x$ $-a x^{2}+(2-a) x$. 若 $y=f(x)$ 的图象与 $x$ 轴交于 $A, B$ 两点, 线段$AB$中点的横坐标为 $x_{0},$ 证明: $f^{\prime}\left(x_{0}\right)$ $<0$.
解析
由 $f\left(x_{1}\right)=f\left(x_{2}\right)=0,$ 知 $\ln x_{1}-a x_{1}^{2}$ $+(2-a) x_{1}=\ln x_{2}-a x_{2}^{2}+(2-a) x_{2},$
故$a=\dfrac{\ln x_{1}-\ln x_{2}+2\left(x_{1}-x_{2}\right)}{x_{1}^{2}-x_{2}^{2}+x_{1}-x_{2}}$
$f^{\prime}\left(x_{0}\right) < 0$
$\Leftrightarrow x_{0}=\dfrac{x_{1}+x_{2}}{2}>\dfrac{1}{a}$
$\Leftrightarrow \dfrac{x_{1}+x_{2}}{2}>\dfrac{x_{1}^{2}+x_{2}^{2}+x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}+2\left(x_{1}-x_{2}\right)}$
$\Leftrightarrow \dfrac{x_{1}+x_{2}}{2}>\dfrac{x_{1}+x_{2}+1}{\frac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}+2}$
$\Leftrightarrow \dfrac{2}{x_{1}+x_{2}}<\dfrac{\ln x_{1}-\ln x_{2}}{x_{1}-x_{2}}$

总结

$\color{grey}{b>\sqrt{\dfrac{a^{2}+b^{2}}{2}}>}$$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}>\sqrt{a b}$$\color{grey}{>\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}>a}$

题号/来源	所用的不等式	方程组处理
1/2016国一21	$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}$	相减、反证法
2	$\dfrac{2}{a+b}<\dfrac{\ln a-\ln b}{a-b}$	相减
3/2014南通二模20	$\dfrac{b-a}{\ln b-\ln a}>\sqrt{a b}$ $\dfrac{a+b}{2}>\sqrt{a b}$	取对、相减、相加
4/2014陕西21	$\color{grey}{ b>\dfrac{b-a}{\ln b-\ln a}}$或$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}$	命$b=n+1$, $a=n$
5/2013全国大纲卷22II	$\dfrac{b-a}{\ln b-\ln a}>\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}$	裂项
6/2010天津理21	$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}$	相减、取对
7/2018全国I卷21	$\dfrac{\ln b-\ln a}{b-a}<\dfrac{1}{\sqrt{ab}}$	相减、韦达
8/2016湖南高联预15	$\dfrac{a+b}{2}>\dfrac{b-a}{\ln b-\ln a}$	相减
9/2018福建高联预14	$\dfrac{2}{a+b}<\dfrac{\ln a-\ln b}{a-b}$	取对、相减
10

参考资料

《利用对数均值不等式破解极值点偏移问题》陈友镇
《巧用对数均值不等式解高考压轴题》彭耿铃
《对数均值不等式在高考及竞赛中的运用》陈纪刚
《妙用对数均值不等式解题》徐春艳

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