对数均值不等式在高考题中的运用

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

1. 证明ALG不等式；
2. 根据\(f(x_1)=f(x_2)=a\)联立2个等式；
3. 指数化对数，有\(a\)的相减消\(a\)；
4. 得到关系，代入具体题目分析。
5. *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

1. 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 \(x_{1} \cdot x_{2}>e^{2}\).

   设2个零点为\(x_{1}, x_{2}\). 则有

   \(\left\{\begin{array}{l}\ln x_{1}=a x_{1} \\ \ln x_{2}=a x_{2}\end{array}\right.\)

   \(\ln x_{1}-\ln x_{2}=a\left(x_{1}-x_{2}\right)\)

   \(\dfrac{1}{a}=\dfrac{x_{1}-x_{2}}{\ln x_{1}-\ln x_{2}}<\dfrac{x_{1}+x_{2}}{2}\)

   \(\therefore a >x_{1} \dfrac{2}{x_{1}+x_{2}}\)

   \(\ln x_{1}+\ln x_{2}=a\left(x_{1}+x_{2}\right)\) \(>\dfrac{2}{x_{1}+x_{2}} \left(x_{1}+x_{2}\right)=2\)

   \(\therefore\ln \left(x_{1} x_{2}\right)>2\), \(x_{1} x_{2}>e^{2}\)

2. 证明: \(x-\ln x>2 \dfrac{e^{x}-x}{e^{x}+x}\).

   \(\dfrac{\ln e^{x}-\ln x}{e^{x}-x}>\dfrac{2}{e^{x}+x}\)

   \(x-\ln x>2 \dfrac{e^{x}-x}{e^{x}+x}\)

3. 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.\(\left[\dfrac{1}{2},+\infty\right)\)

   D. 解析：$x>1 , x<(x-) a(x-) $ \(\Rightarrow a \geqslant \dfrac{1}{2}\)

4. 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.\(\left[\dfrac{1}{2},+\infty\right)\)

   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

1. (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 : (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 ( 1- x _1 ) -2 ( x _2-1 ) = ( 2- x 1 )- (2-x{2})- $,

   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 \(, 则\) 1-x _1 x _2-1 $,

   所以$ $,

   所以$1 < $,

   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 $.

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: \(f'\left(\dfrac{x_{1}+x_{2}}{2}\right)<0\)

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 \(\left(\dfrac{-a+\sqrt{a^{2}+16}}{4},+\infty\right)\)

2. 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(x_{1}-x_{2} ) -(x_{1}^{2}-x_{2}{2})=a(x_{1}-x_{2}) $

\(\Leftrightarrow\dfrac{2\left(\ln x_{1}-\ln x_{2}\right)-\left(x_{1}^{2}-x_{2}^{2}\right)}{x_{1}-x_{2}}=a\),

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.

3. (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(x_{2}, 0)$ 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：(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.

4. （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.

   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】这里如果用$b> $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：

   \(\dfrac{(n+1)-n}{\ln(n+1)-\ln n}<\dfrac{n+1+n}{2}\) \(\dfrac{1}{\ln (n+1)-\ln n}<\dfrac{2 n+1}{2}\) \(\ln (n+1)-\ln n>\dfrac{2}{2 n+1}>\dfrac{2}{2 n+2}=\dfrac{1}{n+1}\)

5. （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 \(: a_{2 n}-a_{n}+\dfrac{1}{4 n}>\ln 2\).

   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 \(a_{2 n}-a_{n}+\dfrac{1}{4 n}>\ln 2\)

6. (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\(x_{1}+\) \(x_{2}>2\).

   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\(x_{1}+x_{2}>2\).

7. (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: \(\dfrac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}<a-2\).

   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.

8. (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: \(x_{1} x_{2}>e^{2}\).

   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: $(x_{1} x_{2})>m =2 $ \(\Rightarrow x_{1} x_{2}>e^{2}\)成立。

9. (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:\(x_{1}+x_{2}>2\).

   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.

10. 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: \(f^{\prime}\left(x_{0}\right)\) \(<0\).

    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