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\(x_{1}, x_{2}\).
- 求a的取值范围; (答案：\(a \in\left(0,\frac{1}{e}\right)\))
- 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 \(f(x)=2 \ln x-x^{2}-a x\).
求函数 \(f(x)\) 的单调区间;
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 \(\left(\dfrac{-a+\sqrt{a^{2}+16}}{4},+\infty\right)\)
- 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 \(x_{1}<x_{2}\).
求实数a 的取值范围;
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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