$f(x)=\cfrac{x}{e^x} ,若f(x_1)=f(x_2),求证x_1+x_2\gt 2.及x_1x_2\lt 1$

$证明：不难知道x_1,x_2\gt 0$
$①x_1,x_2捆绑，不自由②地位相同，因此可用"不妨设"$
$不妨设0\lt x_1\lt x_2$
$①若x_2\in [2,+\infty),x_1\gt 0$
$\Rightarrow x_1+x_2\gt 2 不证自明$
$②{f}' (x)=e^{-x}(1-x)\Rightarrow f(x)在(-\infty,1)\nearrow ;(1,+\infty)\searrow $
$\therefore 0\lt x_1\lt 1\lt x_2$
$当x_2\in (1,2)，要证x_1+x_2\gt 2即证x_1\gt 2-x_2(两边都要有未知数）$
$即证{\color{Red} f(x_1)\gt f(2-x_2),这一步加上f套子，下一步消元} $
$f(x_2)\gt f(2-x_2)\quad 变成一元问题。$
$令F(x)=f(x)- f(2-x)\quad x\in (1,2)$
$证明F(x)\gt 0,\quad 先考虑端点$
$F(1)=f(1)-f(1)=0,{\color{Red}对称构造嘛 } 即证F(x)在(1,2)单调递增$
${F}' (x)={f}' (x)-{(2-x)}' {f}' (2-x)=\cfrac{1-x}{e^x} +\cfrac{1-(2-x)}{e^{2-x}} $
$=(x-1)(\cfrac{1}{e^{2-x}} -\cfrac{1}{e^x} )$
$\because x\gt 2-x,\therefore {F}' (x)\gt 0$
$\therefore {F} (x)\nearrow ,F(x0\gt F(1)=0原式成立。$
以上是方法——，对称构造函数法；下面是方法二，换元法。
$证明：f(x_1)=f(x_2)=a\Rightarrow \begin{cases} \cfrac{x_1}{e^{x_1}}=a\\ \cfrac{x_2}{e^{x_2}}=a\end{cases}\Rightarrow \begin{cases} x_1=ae^{x_1}\quad ① \\x_2=ae^{x_2}\quad ② \end{cases}$
${\color{Green} 换元要有中间过渡的变量，题目中没有，我们就自己设a参量} $
$① -② x_1-x_2=a(e^{x_1}-e^{x_2})\Rightarrow a=\cfrac{x_1-x_2}{e^{x_1}-e^{x_2}} $
$要证x_1+x_2\gt 2即证a(e^{x_1}+e^{x_2})\gt 2$
$a(e^{x_1}+e^{x_2})=\cfrac{x_1-x_2}{e^{x_1}-e^{x_2}} (e^{x_1}+e^{x_2})=\cfrac{e^{x_1}+e^{x_2}}{e^{x_1}-e^{x_2}} (x_1-x_2)$
$这时已经是齐次式了，{\color{Red} \Rightarrow \begin{cases} \ln 换元经常用到t=\cfrac{x_1}{x_2}\\ e^{x} 换元要用到t=x_1-x_2\end{cases} } $
$\cfrac{e^{x_1}+e^{x_2}}{e^{x_1}-e^{x_2}} (x_1-x_2)=(x_x-x_2)\cdot \cfrac{e^{x_1-x_2}+1}{e^{x_1-x_2}-1}$
$不妨设x_1\gt x_2,x_1-x_2\in (0,+\infty),令t=x_1-x_2 $
$即证t\cdot \cfrac{e^t+1}{e^t-1}\gt 2,t\gt 0,e^t\gt 1$
$即证t\cdot (e^t+1)\gt 2(e^t-1),$
$即证(t-2) e^t+t+2\gt 0 \quad t\in (0,+\infty),$
$g(t)=(t-2) e^t+t+2,$
$g(0)=-2\times 1=2=0$
${g}'(t)=e^t(1+t-2)+1=e^t(t-1), {g}'(0)=0$
$\varphi (x)={g}'(t)=e^t(t-1)+1 \Rightarrow {\varphi }' (x)=te^et\gt 0$
$\Rightarrow {g}'(t)\nearrow {g}'(t)\gt {g}'(0)=0\Rightarrow g(t)\nearrow $
$g(t)\gt g(0)=0得证$
$总结：引入中间参量a,做减法，齐次化，ln 换元用\frac{x_1}{x_2},e^x换元用x_1-x_2 $
方法三：对数均值不等式ALG
$f(x_1)=f(x_2)\Rightarrow 不设参量a ,\quad \cfrac{x_1}{e^{x_1}}=\cfrac{x_2}{e^{x_2}}$
$两边取对数,\Rightarrow \ln x_1-x_1=\ln x_2-x_2\Rightarrow x_1-x_2=\ln x_1-\ln x_2$
$\Rightarrow {\color{Red} \cfrac{x_1-x_2}{\ln x_1-\ln x_2}=1,}$
$要证x_1+x_2\gt 2\Rightarrow \cfrac{x_1+x_2}{2}\gt 1$
$即证 \cfrac{x_1+x_2}{2}\gt \cfrac{x_1-x_2}{\ln x_1-\ln x_2}$
$g(a_0)=a_0+{\color{Orange} \frac{1}{a_0} } -a_0\times{\color{Orange} \frac{1}{a_0} }-2 =a_0+{\color{Orange} \frac{1}{a_0} } -3$

$要证x_1x_2\lt 1，也可以用对数均值不等式$
$法一：用对数均值不等式，从上述操作已得到\cfrac{x_1-x_2}{\ln x_1-\ln x_2}=1,它是比几何平均数要大的。略。$

以下为5-7号添加
法二：对称构造辅助函数法
$不妨设0\lt x_1\lt 1\lt x_2欲证x_1x_2\lt 1\Leftrightarrow x_1\lt \cfrac{1}{x_2} \lt 1$
$即证f(x_1)\lt f(\cfrac{1}{x_2}) \Leftrightarrow {\color{Green} f(x_2)\lt f(\cfrac{1}{x_2})} $
$构造辅助函数F(x)=f(x)-f(\cfrac{1}{x})\quad (x\gt 1)$
$F(1)=f(1)-f(1)=0,即证F(x)在x\in (1,+\infty)单调递减。$
${F}' (x)={f}' (x)-{(\cfrac{1}{x})}'{f}' (\cfrac{1}{x}) =\cfrac{1-x}{e^x} +\cfrac{1}{x^2} \cfrac{1-\cfrac{1}{x} }{e^{\cfrac{1}{x} }} $
${F}' (x)=\cfrac{1-x}{e^x} +\cfrac{x-1}{x^3 e^\cfrac{1}{x} }=(x-1)(\cfrac{1}{x^3 e^\cfrac{1}{x} } -\cfrac{1}{e^x}) $
$不容易判断分母的大小。$
$应注意{\color{Green} f(x_2)\lt f(\cfrac{1}{x_2})} 即证\Leftrightarrow{\color{Red} \cfrac{x_2}{e^{x_2}} \lt \cfrac{\frac{1}{x_2} }{e^{\frac{1}{x_2} }} }$
$\Leftrightarrow {\color{Orange} \ln } (\cfrac{x_2}{e^{x_2}} )\lt {\color{Orange} \ln} (\cfrac{\frac{1}{x_2} }{e^{\frac{1}{x_2} }} )$
$\Leftrightarrow \ln x_2-x_2 \lt \ln \cfrac{1}{x_2} -\frac{1}{x_2} \Leftrightarrow \ln x_2+\ln x_2\lt x_2-\cfrac{1}{x_2}$
变成了飘带不等式
练习一题：
$已知函数f(x)=4x-\cfrac{1}{2}x^2 -a \ln x(a\gt 0)$
$(1)当a=3时，讨论函数f(x)的单调性。$
$(2)若f(x)有两个极值点x_1,x_2(x_1\lt x_2)$
$求a的取值范围。证明:f(x_1)+f(x_2)\lt 10-\ln a$
$解：定义域：x\in (0,+\infty)$
$(1){f}' (x)=4-x-\cfrac{a}{x} =\cfrac{-x^2+4x-a}{x}$
$当a=3时，{f}' (x)=\cfrac{-x^2+4x-3}{x} =\cfrac{(-x+1)(x-3)}{x}$
$故x\in (0,1)时，{f}' (x)\lt 0,f(x)\searrow ;$
$x\in (1,3)时，{f}' (x)\gt 0,f(x)\nearrow ;$
$x\in (3,+\infty)时，{f}' (x)\lt 0,f(x)\searrow$
$\therefore f(x)在x=1处有极小值f(1);x=3处有极大值f(3)。$
$(2)\quad {f}' (x)=\cfrac{-x^2+4x-a}{x}\quad a\gt 0,有两个零点为x_1,x_2;$
$\Rightarrow\bigtriangleup =4^2-4\times(-1)\times (-a)=16-4a\gt 0 $
${\color{Green}\Rightarrow0\lt a\lt 4 }$
$\Rightarrow {\color{Red} \begin{cases}x_1+x_2=4\\ x_1x_2=a \end{cases}} $
$f(x_1)+f(x_2)=4(x_1+x_2)-\cfrac{1}{2} (x_1^2+x_2^2)-a(\ln x_1+\ln x_2)$
$=4\times {\color{Red} 4} -\cfrac{1}{2} ({\color{Red} 4^2-2\times a} )-a{\color{Red} \ln a}$
$=8+a-a{\color{Red} \ln a} $
$即证8+a-a{\color{Red} \ln a} \lt 10-\ln a$
$即证0\lt a\lt 4时a+\ln a-a\ln a-2\lt 0恒成立,$
${\color{Red} 即证(a+\ln a-a\ln a-2)_{max}} \lt 0$
$设g(a)=a+\ln a-a\ln a-2$
${g}' (a)=1+\cfrac{1}{a} -(\ln a+1)=\cfrac{1}{a} -\ln a,\searrow $
${g}'(1)=1,{g}'(2)=\cfrac{1}{2}-\ln 2\lt 0 $
$\therefore \exists a_0\in (1,2), 使得{g}' (a_0)=0{\color{Red} \Rightarrow \ln a_0=\cfrac{1}{a_0} } $
$故a\in (0,a_0),{g}' (a)\gt 0,g(a)\nearrow ;$
$a\in (a_0,4),{g}' (a)\lt 0,g(a)\searrow ;$
$g(a)在a_0处有极大值g(a_0)=a_0+\ln a_0-a_0\ln a_0-2$
$g(a_0)=a_0+{\color{Orange} \cfrac{1}{a_0} } -a_0\times{\color{Orange} \cfrac{1}{a_0} }-2 =a_0+{\color{Orange} \cfrac{1}{a_0} } -3$
${\color{Orange} \because a_0\in (1,2),容易得到g(a_0)\in (-1,-\cfrac{1}{2} )} ,\quad \therefore g(a_0)\lt 0$

$例3、f(x)=e^x-ax+a图像与x轴交于A(x_1,0)B(x_2,0)两点，求证：x_1x_2\lt x_1+x_2.$
解：$f(x)=0\Rightarrow e^x=ax-a\Rightarrow \begin{cases} e^{x_1}=a(x_1-1)\qquad ①\\e^{x_2}=a(x_2-1)\qquad ②\end{cases}$
$\Rightarrow ①{\div} ②得\Rightarrow \cfrac{e^{\color{Red} {x_1}} }{e^{\color{Red} {x_2}} } =\cfrac{ x_1-1 }{ x_2-1} 两边取对数得$
$\Rightarrow \cfrac{\ln e^{\color{Red} {x_1-1}} }{\ln e^{\color{Red} {x_2-1}} } =\cfrac{ \ln (x_1-1 )}{ \ln (x_2-1)}这是错误的操作，错在哪里？ $
$\Rightarrow (x_1-{\color{Red} 1} )-(x_2-{\color{Red} 1} )=\ln( x_1-1)-\ln ( x_2-1){\color{Red} \Rightarrow \cfrac{(x_1-1)-(x_2-1)}{\ln( x_1-1)-\ln ( x_2-1)} =1\gt \sqrt{( x_1-1)( x_2-1)} } $
$( x_1-1)( x_2-1)\lt 1{\color{Green} \Rightarrow } x_1x_2-(x_1+x_2)+1\lt 1{\color{Peach} \Rightarrow x_1x_2\lt x_1+x_2}$
$当然，应用对数均值不等式要先证后用。严格来说，解题过程还要证明a\gt 0,即证x_1-1\gt 0. $
$由题意有两点，可得{f}' (x)=e^x-a\Rightarrow {f}' (x_0)=0\Rightarrow x_0=\ln a且有f (x_0)\lt 0$
$f(x_0)=e^{x_0}-ax_0+a=e^{\ln a}-a\ln a+a=2a-a\ln a=a(2-\ln a)\lt 0$
$\Rightarrow a\gt e^2$

$例4、f(x)=2\ln x-x^2-mx与x轴交于x_1,x_2,x_1\lt x_2,求证{f}' (\cfrac{1}{3}x_1+\cfrac{2}{3}x_2 )\lt 0$
$分析\begin{cases} 2\ln x_1=x_1^2+mx_1\\2\ln x_2=x_2^2+mx_2 \end{cases},{f}' (x)=\cfrac{2}{x} -2x-m,显然{\color{Red} {f}' (x)\searrow } $
${f}'(x_0)=0\Rightarrow m=\cfrac{2}{x_0}-2x_0 $
$要证{f}' (\cfrac{1}{3}x_1+\cfrac{2}{3}x_2 )\lt 0={f}' (x_0)$
$即证{\color{Red} \cfrac{1}{3}x_1+\cfrac{2}{3}x_2 \gt x_0 }\Leftrightarrow {\color{Red} x_1+2x_2 \gt 3x_0 } $
$\because x_1\lt x_0\lt x_2\Rightarrow x_1+x_2+x_2\gt 2x_0+x_0\Leftrightarrow {\color{Red} x_1+x_2 \gt 2x_0 }$
$解：若有x_1+x_2\gt 2x_0成立，\because x_0是极值点，\therefore x_1\lt x_0\lt x_2\quad \therefore x_2\gt x_0 $
${\color{Red} \Rightarrow x_1+2x_2\gt 3x_0\Rightarrow \cfrac{1}{3}x_1+\cfrac{2}{3}x_2 \gt x_0 } $
$①若x_2\in [2x_0,+\infty),x_1\gt 0,\Rightarrow x_1+x_2\gt x_2\ge 2x_0{\color{Red} \Rightarrow x_1+x_2\gt 2x_0}$
$②若x_2\in (x_0,2x_0),x_1\in (0,x_0),{\color{Green} x_1\gt 2x_0-x_2}$
${f}' (x)=\cfrac{2}{x}-2x-m\quad (x\gt 0){\color{Red} \searrow x_0是导函数的零点，}$
${\color{Red} m=\cfrac{2}{x_0}-2x_0} ,虚设零点$
$x\in (0,x_0),{f}' (x)\gt 0={f}' (x_0),f(x)\nearrow$
$x\in (x_0,+\infty),{f}' (x)\lt 0={f}' (x_0),f(x)\searrow$
$x_2\in (x_0,2x_0),2x_0-x_2\in (0,x_0)$
$即证:f(x_1)\gt f(2x_0-x_2)$
$\because f(x_1)=f(x_2),即证f(x_2)\gt f(2x_0-x_2)\quad (消元)$
$即证{\color{Orange} F(x)=f(x)- f(2x_0-x)\gt 0\quad x\in (x_0,2x_0)} $
${\color{Orange} F(x_0)=f(x_0)- f(2x_0-x_0)=0},{\color{Red} 即证F(x)在x\in (x_0,2x_0)\nearrow } $
${F}' (x)={f}' (x)+{f}' (2x_0-x)=(\cfrac{2}{x}-2x-m )+[\cfrac{2}{2x_0-x}-2(2x_0-x)-m ]$
$=\cfrac{2}{x}+\cfrac{2}{2x_0-x}-4x_0-2(\cfrac{2}{x_0} -2x_0)$
$=\cfrac{2}{x}+\cfrac{2}{2x_0-x}-\cfrac{4}{x_0} $
$=\cfrac{2}{x}-\cfrac{2}{x_0} +\cfrac{2}{2x_0-x}-\cfrac{2}{x_0} $
$=2(\cfrac{1}{x}-\cfrac{1}{x_0}) +2(\cfrac{1}{2x_0-x}-\cfrac{1}{x_0} )$
$=2\cfrac{x_0-x}{xx_0} +2\cfrac{x_0-(2x_0-x)}{x_0(2x_0-x)}$
$=2\cfrac{x_0-x}{xx_0} -2\cfrac{x_0-x}{x_0(2x_0-x)}$
$=2\cfrac{(x_0-x)}{x_0} [\cfrac{1}{x} -\cfrac{1}{2x_0-x}]$
$=2\cfrac{(x_0-x)}{x_0} \cfrac{(2x_0-x-x)}{x(2x_0-x)}$
$=\cfrac{4(x_0-x)^2}{x_0x(2x_0-x)}\gt 0$
${F}' (x)\gt 0,\therefore F(x)在x\in (x_0,2x_0)是单增的。F(x)\gt F(x_0)=0$
以上方法是进行了放缩，先变成对称变量，再构造。
其实我们直接构造辅助函数也能证明，只是判断辅助函数的导函数的正负过程计算量不少。
${\color{Red} 要证} {f}' (\cfrac{1}{3}x_1+\cfrac{2}{3}x_2)\lt {f}'(x_0) {\color{Red}\qquad \because {f}'(x)\searrow } $
${\color{Red} 即证} \cfrac{1}{3}x_1+\cfrac{2}{3}x_2\gt x_0$
${\color{Red} 即证} x_1+2x_2\gt 3x_0,\qquad x_2\gt x_1$
$①若x_2\in [\cfrac{3}{2}x_0,+\infty),2x_2\ge 3x_0,x_1\gt 0 $
$\Rightarrow x_1+2x_2\gt 3x_0\quad (不证自明)$
$②若x_2\in (x_0,\cfrac{3}{2}x_0),$
${\color{Red} 即证}x_1\gt 3x_0-2x_2\quad 检查左右两边是否属于同一单调区间。$
$\because {f}' (x)在(0,+\infty)\searrow $
$\therefore x\in (0,x_0),{f}' (x)\gt {f}' (x_0)=0$
$x\in (x_0,+\infty),{f}' (x)\lt 0$
$\therefore f(x)在(0,x_0)\nearrow ;在(x_0,+\infty)\searrow $
${\color{Red} 即证}：f(x_1)\gt f(3x_0-2x_2)$
${\color{Red} 即证}：f(x_2)\gt f(3x_0-2x_2)$
${\color{Red} 设F(x)}=f(x)-f(3x_0-2x)\gt 0在x\in (x_0,\cfrac{3}{2}x_0)成立；$
${\color{Red} F(x_0)}=f(x_0)-f(3x_0-2x_0)=0$
${\color{Red} {F}' (x)}={f}' (x){\color{Green} +2} {f}' (3x-2x)=(\cfrac{2}{x}-2x-m )+2[\cfrac{2}{3x_0-2x} -2(3x_0-2x)-m]$
$=\cfrac{2}{x}+\cfrac{4}{3x_0-2x} +6x-12x-3(\cfrac{2}{x_0}-2x_0)$
$= \cfrac{2}{x}+\cfrac{4}{3x_0-2x} -\cfrac{6}{x_0}+6x-6x_0$
$= \cfrac{2}{x}-\cfrac{2}{x_0}+\cfrac{4}{3x_0-2x} -\cfrac{4}{x_0}+6(x-x_0) $
$=\cfrac{2(x_0-x)}{x_0x} +\cfrac{8(x-x_0)}{x_0(3x_0-2x)} +6(x-x_0)$
$=\cfrac{2(x-x_0)}{x_0}(-\cfrac{1}{x} +\cfrac{4}{3x_0-2x} ) +6(x-x_0)$
$=\cfrac{2(x-x_0)}{x_0}\times\cfrac{2x-3x_0+4x}{x(3x_0-2x)} +6(x-x_0)$
$\because x\gt x_0\quad \therefore 6x\gt 3x_0,x-x_0\gt 0,$
$\Rightarrow {F }'(x)\gt 0,F(x)在(x_0,\cfrac{3}{2} x_0)\nearrow ,F(x)\gt F(x_0)=0，得证$

换元法：

$例5、f(x)=-\cfrac{a}{2}e^{2x}+(x-1)e^x有两个极值点x_1,x_2,且x_1\lt x_2,求证:x_1+2x_2\gt 3$
$解：{f}' (x)=-ae^{2x}+(x-1+1)e^x=-ae^{2x}+xe^x=0\Rightarrow -ae^x+x=0$
$x=ae^x\Rightarrow \begin{cases} x_1=ae^{x_1}\quad ①\\x_2=ae^{x_2}\quad ②\end{cases}$
$ ①- ②,得a=\cfrac{x_1-x_2}{e^{x_1}-e^{x_2}}$
$x_1+2x_2=a(e^{x_1}+2e^{x_2})=\cfrac{x_1-x_2}{e^{x_1}-e^{x_2}} (e^{x_1}+2e^{x_2})$
$=\cfrac{(e^{x_1}+2e^{x_2})}{e^{x_1}-e^{x_2}} (x_1-x_2)=\cfrac{e^{x_1-x_2}+2}{e^{x_1-x_2}-1} (x_1-x_2)\gt 3$
${\color{Red} 要证} x_1+2x_2\gt 3,设t=x_1-x_2,\quad (t\lt 0,e^t\lt 0)$
${\color{Red} 即证} \cfrac{e^t+2}{e^t-1} t \gt 3\quad ({\color{Red} t\lt 0,e^t-1\lt 0})$
${\color{Red} 即证} (e^t+2)t \lt 3(e^t-1)\quad t\in(-\infty,0)$
$令\varphi (x)=(e^t+2)t - 3(e^t-1)=e^t(t-3)+2t+3,\quad t\in(-\infty,0)$
${\color{Red} 即证\varphi (x)_{max}\lt 0 ,\quad t\in(-\infty,0)}$
$\varphi (0)=e^0(0-3)+20+3=0$
${\varphi }' (x)=e^t(t-3+1)+2=e^t(t-2)+2$
${\varphi }' (0)=e^0(0-2)+2=0$
${\varphi }'' (t)=e^t(t-1)\lt 0\Rightarrow {\varphi }' (t)\searrow 且{\varphi }' (0)=0,\Rightarrow {\varphi }' (t)\gt 0\quad t\lt 0$
$\Rightarrow \varphi (t)\nearrow ,\varphi (t)\lt \varphi (0)=0,得证$

https://one.free.nf/admin/write-post.php?cid=197 例2
$例6.已知函数f(x)=\ln x -ax^2+(2-a)x与x轴交于AB两点，AB中点横坐标为x_0，$
$求证:{f}' (x_0)\lt 0$
$解： {\color{Red} 定义域为(0,+\infty) } $
${f}' (x)=\cfrac{1}{x}-2ax+2-a=\cfrac{-2ax^2+(2-a)x+1}{x}$
$=\cfrac{(2x+1)(-ax+1)}{x} =\cfrac{2x+1}{x} \cdot (-ax+1)$
$(1)a\in (-\infty,0],-ax\ge 0,-ax+1\gt 0$
$\therefore {f}' (x)\gt 0,f(x)在(0,+\infty)\nearrow ,舍去。$
$(2)a\in (0,+\infty),令{f}' (x)\gt 0,x\in (0,\cfrac{1}{a} );$
$令{f}' (x)\lt 0,x\in (\cfrac{1}{a},+\infty )$
${\color{Red} \therefore } \quad f(x)在(0,\cfrac{1}{a} )\nearrow ； (\cfrac{1}{a},+\infty )\searrow $
$要满足题意，f(\cfrac{1}{a})\gt 0\Rightarrow \ln \cfrac{1}{a}+\cfrac{1}{a}-1\gt 0\Rightarrow a\in (0,1) $
$要证{f}' (x_0)\lt 0={\color{Red} {f}' (\cfrac{1}{a} )} $
$\because a\in (0,1)\Rightarrow {f}' (x)在(0,+\infty)\searrow ,即证x_0\gt \cfrac{1}{a} {\color{Red} 成功脱掉外套f'} $
$设A(x_1,0),B(x_2,0),即证x_1+x_2\gt \cfrac{2}{a}$
$不妨设x_1\gt x_2,下面是常规的对称构造函数证明极值点偏移。$
$(1)x_2\in [\cfrac{2}{a} ,+\infty),x_1\gt 0,\Rightarrow x_1+x_2\gt \cfrac{2}{a}$
$不证自明$
$(2)x_2\in ( \cfrac{1}{a},\cfrac{2}{a}),{\color{Green} 即证x_1\gt \cfrac{2}{a}-x_2} $
${\color{Green} 即证f(x_1)\gt f(\cfrac{2}{a}-x_2)} $
${\color{Green} 即证f(x_2)\gt f(\cfrac{2}{a}-x_2) } $
$令F(x)=f(x)-f(\cfrac{2}{a}-x) \quad x\in ( \cfrac{1}{a},\cfrac{2}{a})$
$F(\cfrac{1}{a} )=f(\cfrac{1}{a} )-f(\cfrac{2}{a}-\cfrac{1}{a})=0{\color{Red} 即证F(x)\nearrow } $
${F}' (x)={f}' (x){\color{Red} +} {f}'(\cfrac{2 }{a} -x) =\cfrac{2x+1}{x} (-ax+1)+\cfrac{\cfrac{4}{a}-2x+1 }{\cfrac{2 }{a} -x}(ax-1)$
$=(ax-1) (\cfrac{\cfrac{4}{a}-2x+1 }{\cfrac{2 }{a} -x}-\cfrac{2x+1}{x} )$
$=(ax-1) (\cfrac{1}{\cfrac{2 }{a} -x}-\cfrac{1}{x} )$
$\because x\gt \cfrac{1}{a} ax-1\gt0,2x\gt \cfrac{2}{a}\Rightarrow x\gt \cfrac{2}{a}-x\gt 0$
$\Rightarrow \cfrac{1}{x} \lt \cfrac{1}{\cfrac{2}{a}-x} \Rightarrow {F}' (x)\gt 0\Rightarrow F(x)在( \cfrac{1}{a},\cfrac{2}{a})\nearrow$
${\color{Red} \therefore \quad } F(x)\gt F(\cfrac{1}{a} )=0$

非x类
$例7.已知函数f(x)=x\ln x -\cfrac{a}{2} x^2-x有两个极值点x_1,x_2，$
$求证:\cfrac{1}{\ln x_1} +\cfrac{1}{\ln x_2}\gt 2$
$解:定义域为(0,+\infty)$
${f}' (x)=\ln x+1-ax-1=\ln x-ax$
${f}' (x)=0\Rightarrow \ln x=ax有两根x_1,x_2$
$\Rightarrow \begin{cases} ax_1=\ln x_1\\ax_2=\ln x_2\end{cases}\Rightarrow {\color{Green}a=\cfrac{\ln x_1-\ln x_2}{x_1- x_2} } $
$\cfrac{1}{\ln x_1} +\cfrac{1}{\ln x_2} =\cfrac{1}{ax_1} +\cfrac{1}{ ax_2}=\cfrac{1}{a}(\cfrac{1}{x_1} +\cfrac{1}{x_2})$
$=\cfrac{x_1- x_2}{\ln x_1-\ln x_2}(\cfrac{1}{x_1} +\cfrac{1}{x_2})=\cfrac{1}{\ln \cfrac{x_1}{x_2} }(1+\cfrac{x_1}{x_2}-\cfrac{x_2}{x_1}-1)$
$不妨设x_1\gt x_2\gt 0,t=\cfrac{x_1}{x_2} ,t\in (1,+\infty)$
$即证\cfrac{1}{\ln t} (t-\cfrac{1}{t} )\gt 2$
$即证\ln t \lt \cfrac{1}{2} (t-\cfrac{1}{t})\quad t\gt 1$
${\color{Red} 方法二} $
我们还可以使用对称构造辅助函数证明它！
${\color{Red} {f}' (x)=\ln x-ax} \quad {f}'(x)=0有两根x_1,x_2$
$证明：\cfrac{1}{\ln x_1} +\cfrac{1}{\ln x_2}\gt 2，$
$显然a\gt 0,即证\cfrac{1}{x_1}+ \cfrac{1}{x_2}\gt 2a$
$令t_1=\cfrac{1}{x_1}, t_2=\cfrac{1}{x_2}\Rightarrow x_1=\cfrac{1}{t_1}, x_2=\cfrac{1}{t_2}$
$\ln x=ax\Rightarrow \ln \cfrac{1}{t} =\cfrac{a}{t} \Rightarrow {\color{Red} -\ln t=\cfrac{a}{t} } $
$设g(t)=\ln t+\cfrac{a}{t}有两个交点(t_1,0)，(t_2,0)$
${g}'(t)=\cfrac{1}{t}-\cfrac{a}{t^2} =\cfrac{t-a}{t^2} $
$g(t)在(0,a)\searrow ,(a,+\infty)\nearrow $
$不妨设t_1\lt t_2\Rightarrow 0\lt t_1\lt a\lt t_2$
$①若t_2\in [2a,+\infty),t_1\gt 0\Rightarrow t_1+t_2\gt 2a(不证自明)$
$②t_2\in (a,2a),2a-t_2\in (0,a)$
$即证t_1\gt 2a-t_2,即证g(t_1)\lt g(2a-t_2)\quad 套上g$
$即证g(t_2)\lt g(2a-t_2)\quad 此处消元$
$令G(t)=g(t)-g(2a-t)\quad t\in (a,2a)$
$G(a)=g(a)-g(2a-a)=0\quad 即证G(t)\searrow ,心中有数$
${G}' (t)={g}'(t){\color{Red} + } {g}' (2a-t)=\cfrac{t-a}{t^2}+\cfrac{2a-t-a}{(2a-t)^2}$
$=(t-a)[\cfrac{1 }{t^2} -\cfrac{1}{(2a-t)^2}]$
$\because a\lt t,2a\lt 2t{\color{Red} \Rightarrow 2a-t\lt t }$
$\therefore {G}' (t)\lt 0\Rightarrow G(t)在(a,2a)单调递减，$
$G(t)\lt G(a)=0\Rightarrow G(t)\lt 0,得证$

一、主元法：

未知数的关系

$(1)题目中显现，+－\times \div$
$(2)未知数之间的控制关系：
$\begin{cases} ①自由\\ ②捆绑，f(x)=\cfrac{\ln x}{x}=a,此时x_1,x_2捆绑关系 \end{cases}$
$\begin{cases} \quad①地位相同 \\ \quad \\ \quad ②地位不相同2x_1+x_2与x_1+2x_2\end{cases}$

$证明对数不等式ALG:\sqrt{x_1x_2} \lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2} \lt\cfrac{x_1+x_2}{2}$
未知数关系：(1)自由、(2)地位相同
$先证左侧不等式：不妨设：x_1\gt x_2\Rightarrow \ln x_1\gt \ln x_2\Leftrightarrow \ln x_1- \ln x_2\lt \cfrac{x_1-x_2}{\sqrt{x_1x_2} }$
此时可换元，但我们考虑新方法，叫做主元法：
${\color{Red}把x_1视为主元，即自变量，x_2看作是常数、参数 }$
$x_1\in(x_2,+\infty)\quad \because x_1\gt x_2$
$即证：f(x_1)=\ln x_1-\ln x_2-\cfrac{x_1}{\sqrt{x_1} } +\cfrac{x_2}{\sqrt{x_2} } \Leftrightarrow f(x_1)\lt 0$
${\color{Green} 这里x_1是自变量，x_2是参数，常数} $
$显然有f(x_2)=\ln x_2-\ln x_2-\cfrac{x_2}{\sqrt{x_2} } +\cfrac{x_2}{\sqrt{x_2} } =0$
$这里是把作为常数的x_2代入x_1为自变量的上式中的x_1。$
$只需证f(x_1)在(x_2,+\infty)单减$
${f}' (x_1)=\cfrac{1}{x_1} -\cfrac{1}{\sqrt{x_2}} \cdot \cfrac{1}{2\sqrt{x_1} } +\sqrt{x_2} (-\cfrac{1}{2}\cdot \cfrac{1}{x_1^{ \frac{3}{2}} } )$
$=\cfrac{2\sqrt{x_2}\cdot \sqrt{x_1} -x_1-\sqrt{x_2} \sqrt{x_2} }{2\sqrt{x_2} \cdot x_1^{\frac{3}{2} }}$
$= \cfrac{2\sqrt{x_1x_2}-x_1-x_2 }{2\sqrt{x_2} \cdot x_1^{\frac{3}{2} }}=-\cfrac{(\sqrt{x_1}-\sqrt{x_2})^2 }{2\sqrt{x_2} \cdot x_1^{\frac{3}{2} }}\lt 0$
$\therefore f(x_1)在(x_2,+\infty )单减，\Rightarrow f(x_1)\lt f(x_2)=0$

$例1、已知f(x)=e^x\ln (x+1),其导函数在[0,+\infty)单增，证明：$
$对于\forall s,t\in (0,+\infty),f(s+t)\gt f(s)+f(t)$
分析：s,t自由，地位相同
$解：本题即证：H(x)=f(s+t)-f(s)-f(t)在 (0,+\infty)大于0$
${\color{Red}把s视为主元，即自变量，t看作是常数、参数 }$
$H(0)=f(0+t)-f(0)-f(t)在 (0,+\infty)=0=f(0)=0$
$\because f(0)=e^0\cdot 0=0$
$只需证H(s)在(0,+\infty)单增$
{H}' (s)={f}'(s+t) -{f}'(s)
$又\because s+t\gt s,f(x)导函数在[0,+\infty)单增$
$\Rightarrow {H}' (s)={f}'(s+t) -{f}'(s) \gt 0\Rightarrow$
$H(s)在(0,+\infty)\nearrow ,H(s)\gt H(0)=0,故原式成立。$

二、换元法

${\color{Red} 换元法包括 } \begin{cases} \quad 换函数分析\\\quad\\ \quad 换未知数分析\end{cases}$

换函数分析

$例1、g(x)=\cfrac{1-e^x}{x} ,\forall x_1,x_2\in [1,e],\cfrac{g(x_1)-g(x_2)}{x_1-x_2} \lt \cfrac{m}{x_1x_2}恒成立，求m的取值范围$
$x_1,x_2自由，且地位相同，（主元法不好弄）$
$解:不妨设x_1\gt x_2, x_1-x_2\gt 0\Rightarrow g(x_1)-g(x_2)\lt \cfrac{m(x_1-x_2)}{x_1x_2} =\cfrac{m}{x_2}-\cfrac{m}{x_1}$
$g(x_1)-g(x_2)\lt \cfrac{m}{x_2}-\cfrac{m}{x_1}$
$\Rightarrow g(x_1)+\cfrac{m}{x_1}\lt g(x_2)+\cfrac{m}{x_2}$
$设f(x)=g(x)+\cfrac{m}{x},f(x_1)\lt f(x_2)说明f(x)在[1,e]\searrow$
$即 {f}' (x)={g}' (x)-\cfrac{m}{x^2}\le 0在在[1,e]恒成立$
$\cfrac{e^x-1-xe^x}{x^2} -\cfrac{m}{x^2}\le 0\Rightarrow e^x-1-xe^x\le m$
$令h(x)=(1-x)e^x-1,x\in [1,+\infty),h(x)_{max}\le m$
${h}' (x)=e^x(1-x-1)=-xe^x\lt 0$
$h(x)在[1,e]单减，m\ge h(1)=-1$
$\therefore m\in [-1,\infty)$

$例2、f(x)=x^2+2\ln (x+1)-ax,若m,n\in (-1,+\infty) ,且\cfrac{f(m)-f(n)}{m-n} \gt \cfrac{1}{2} 恒成立，求a的取值范围。$
$解：不妨设m\gt n,f(m)-f(n)\gt \cfrac{1}{2} (m-n)$
$f(m)-\cfrac{1}{2} m\gt f(n)-\cfrac{1}{2} n$
$令g(x)=f(x)-\cfrac{1}{2} x,g(m)\gt g(n)$
$g(x)在[-1,+\infty)\nearrow ,{g}' (x)={f}' (x)-\cfrac{1}{2} \ge 0$
$2x+\cfrac{2}{x+1} -a-\cfrac{1}{2}\ge 0 在(-1,+\infty)恒成立。$
$a\le 2x+\cfrac{2}{x+1} -\cfrac{1}{2}=2(x+1)+\cfrac{2}{x+1} -2-\cfrac{1}{2}$
$2(x+1)+\cfrac{2}{x+1} -2-\cfrac{1}{2}\ge 2\sqrt{4} -\cfrac{5}{2}=\cfrac{3}{2}$
$当2(x+1)=\cfrac{2}{x+1}，即x=0时等号成立，a\in (-\infty,\cfrac{3}{2}]$

换未知数分析

例1:ALG

$例2、a\ge 3已知f(x)=\ln x+x^2-ax有两个极值点x_1,x_2,x_1\lt x_2$
$证明 ：f(x_1)-f(x_2)\ge \cfrac{3}{4} -\ln 2$
$x_1x_2①不自由②地位不相同，因而不能用主元法；$
$解：{f}' (x)=\cfrac{1}{x} +2x-a=\cfrac{2x^2-ax+1}{x}=0\quad有两个实根x_1,x_2,且x_1\lt x_2$
$\Rightarrow 2x^2-ax+1=0,\bigtriangleup =a^2-8\gt 0 \begin{cases} \quad x_1+x_2=\cfrac{a}{2}\\ \quad \\ \quad x_1x_2=\cfrac{1}{2} \end{cases}{\color{Green} \Rightarrow } \begin{cases} \quad {\color{Orange} 2(x_1+x_2)=a} \\ \quad \\ \qquad {\color{Orange}2x_1x_2=1 } \end{cases}$
$f(x_1)-f(x_2)=\ln x_1+x_1^2-ax_1-(\ln x_1+x_2^2-ax_1)=\ln \cfrac{x_1}{x_2}+x_1^2-x_2^2-a(x_1-x_2)$
$分析上式，欲进行齐次换元，只有对数一项是齐次的，后面分别是二次项和一次项。$
$题目中并无条件可利用的条件，除了两个极值点外。$
$观察导函数为零时，由韦达定理得到两个式子的，$
$它们其实就是联系一次项与a,二次项与常数的桥梁$
$\ln \cfrac{x_1}{x_2}+x_1^2-x_2^2-{\color{Orange} a} (x_1-x_2)=\ln \cfrac{x_1}{x_2}+x_1^2-x_2^2-{\color{Orange}2(x_1+x_2) } (x_1-x_2)$
$=\ln \cfrac{x_1}{x_2}-(x_1^2-x_2^2)=\ln \cfrac{x_1}{x_2}-\cfrac{(x_1^2-x_2^2)}{{\color{Orange} 2x_1x_2}}$
$现在已经可以齐次化了，达到了换元的先决条件$
$=\ln \cfrac{x_1}{x_2}-\cfrac{\cfrac{x_1^2}{x_2^2} -1}{\cfrac{x_1}{x_2} } =\ln t-\cfrac{1}{2}(t-\cfrac{1}{t})$
$令t=\cfrac{x_1}{x_2},那么t的取值范围为多少呢？$
$我们目光还要回到韦达定理，题目中a\ge 3,就是说，x_1+x_2\gt {\color{Orange}\cfrac{3}{2} } $
$我们如何从两数和的范围及积为定值得到两数之比的范围？还是齐次化手段！$
$x_1+x_2=\cfrac{a}{2}\ge \cfrac{3}{2} \Rightarrow {\color{Orange} \cfrac{(x_1+x_2)^2}{1}}\ge \cfrac{9}{4} \Rightarrow$
$\Rightarrow \cfrac{(x_1+x_2)^2}{{\color{Orange} 2x_1x_2} } \ge \cfrac{9}{4} \Leftrightarrow \cfrac{(x_1+x_2)^2}{x_1x_2} \ge \cfrac{9}{2} $
$2t-5+\cfrac{2}{t} \ge 0\Rightarrow 2t^2-5t+2 \ge 0\Rightarrow (2t-1)(t-2)\ge 0$
$解得0\lt t \le \cfrac{1}{2}$
$令g(t)=\ln t -\cfrac{1}{2}(t-\cfrac{1}{t})$
$g(\cfrac{1}{2} )=\ln \cfrac{1}{2} -\cfrac{1}{2} (\cfrac{1}{2} -2)=\cfrac{3}{4}-\ln 2$
${g}' (t)=\cfrac{1}{t} -\cfrac{1}{2} (1+\cfrac{1}{t^2})=-\cfrac{(t-1)^2}{t^2}\lt 0,{\color{Red} g(t)\searrow } $
$g(t)\ge g(\cfrac{1}{2} )=\cfrac{3}{4} -\ln 2$
练习：
$f(x)=2-\cfrac{2(a+1)}{x+a} -\ln x,0\lt a \lt 1有两个极值点x_1,x_2,求证：$
$\cfrac{f(x_2)-f(x_1)}{x_2-x_1}\lt \cfrac{1}{a}-\cfrac{2}{a+1}$
解:${f}' (x)=\cfrac{2(a+1)}{(x+a)^2} -\cfrac{1}{x} =\cfrac{2(a+1)x-(x+a)^2}{x(x+a)^2} $
$\cfrac{2(a+1)x-(x^2+2ax+a^2)}{x(x+a)^2} =\cfrac{-x^2+2x-a^2}{x(x+a)^2}$
$\Rightarrow{\color{Green} \begin{cases} x_1+x_2=2\\ x_1x_2=a^2\end{cases}}$
$f(x_2)-f(x_1)=2-\cfrac{2(a+1)}{x_2+a} -\ln x_2-(2-\cfrac{2(a+1)}{x_1+a} -\ln x_1)$
$=\cfrac{2(a+1)}{x_1+a}-\cfrac{2(a+1)}{x_2+a} +\ln x_1-\ln x_2=\ln \cfrac{x_1}{ x_2} +2(a+1)\times \cfrac{x_2+a-(x_1+a)}{(x_1+a)(x_2+a)}$
$=\ln \cfrac{x_1}{ x_2} + \cfrac{2(a+1){\color{Green} (x_2-x_1)} }{(x_1+a)(x_2+a)}$
$\because \cfrac{f(x_2)-f(x_1)}{x_2-x_1}=\cfrac{\ln \cfrac{x_1}{ x_2}}{x_2-x_1} + \cfrac{2(a+1)}{(x_1+a)(x_2+a)} $
$(x_1+a)(x_2+a)={\color{Red} x_1x_2} +a{\color{Red} (x_1+x_2)} +a^2={\color{Red} a^2} +{\color{Red} 2}a+a^2=2a^2+2a$
$\cfrac{f(x_2)-f(x_1)}{x_2-x_1}=-\cfrac{\ln \cfrac{x_1}{ x_2}}{{\color{Green} x_1-x_2} } + \cfrac{2(a+1)}{2a(a+1)}$
$\cfrac{f(x_2)-f(x_1)}{x_2-x_1}=-\cfrac{\ln \cfrac{x_1}{ x_2}}{{\color{Green} x_1-x_2} } + \cfrac{1}{a}\lt \cfrac{1}{a}-\cfrac{2}{a+1}$
${\color{Orange} \Leftrightarrow } \cfrac{\ln \cfrac{x_1}{ x_2}}{{\color{Green} x_1-x_2} } {\color{Red} \gt }\cfrac{2}{a+1}$
$不妨设x_1\gt x_2,\Leftrightarrow \cfrac{\ln \cfrac{x_1}{ x_2}}{{\color{Green} 2} } {\color{Red} \gt }\cfrac{x_1-x_2}{a+1}$
$\because {\color{Green} \begin{cases} x_1+x_2=2\\ x_1x_2=a^2\end{cases}}$
$=\cfrac{x_1-x_2}{a+1}=\cfrac{x_1-x_2}{\sqrt{x_1x_2}+\cfrac{x_1+x_2}{2} } =\cfrac{\cfrac{x_1}{x_2} -1}{\sqrt{\cfrac{x_1}{x_2}} +\cfrac{\cfrac{x_1}{x_2}+1}{2} }$
$令t=\sqrt{\cfrac{x_1}{x_2} } ,t\gt 1$
$\cfrac{1}{2}\ln t^2 \gt \cfrac{t^2 -1}{t+\cfrac{t^2+1}{2} }=\cfrac{2(t^2-1)}{(t+1)^2} =\cfrac{2(t-1)}{t+1} \quad (t\gt 1)$
${\color{Purple} \Leftrightarrow } \ln t \gt \cfrac{2(t-1)}{t+1} $
$7-15增加一题：$
$已知函数f(x)=-ax^2+x-\ln x,a\gt0,若f(x)在定义域内有两个极值点x_1,x_2,求证：f(x_1)+f(x_2)\gt3-\ln 2$
${f}' (x)=-2ax+1-\cfrac{1}{x}=\cfrac{-2ax^2+x-1}{x}$
$\Delta =1^2-4(-2a)(-1)=1-8a\gt 0\Rightarrow a\lt a\lt \cfrac{1}{8}$
${\color{Green} \begin{cases} x_1+x_2=\cfrac{-1}{-2a}=\cfrac{1}{2a} \\x_1x_2=\cfrac{1}{2a} \end{cases}} $
$f(x_1)+f(x_2)=-a(x_1^2+x_2^2)+(x_1+x_2)-\ln x_1-\ln x_2=-a(\cfrac{1}{4a^2} -\cfrac{1}{a})+\cfrac{1}{2a}+\ln 2a$
$\Rightarrow f(x_1)+f(x_2)=1+\cfrac{1 }{4a} +\ln 2a \quad 0\lt a\lt \cfrac{1}{8}$
$设h(a)=1+\cfrac{1 }{4a} +\ln 2a \quad 0\lt a\lt \cfrac{1}{8}$
${h}' (a)=-\cfrac{1}{4a^2}+\cfrac{2}{2a} =\cfrac{4a-1}{4a^2} \lt 0 a\in (0,\cfrac{1}{8} )$
$h(a)\searrow \Rightarrow h(a)\gt h(\cfrac{1}{8} )=3-2\ln 2$

恒成立题型分析
$已知函数f(x)=\cfrac{ax^2+x-1}{e^x}.2018年新3$
$①求曲线y=f(x)在点(0,-1)处的切线方程;$
$②证明:当a\ge 1时,f(x)+e\ge 0\Rightarrow 主元法$
$题型2:已知函数f(x)=(x+1)\ln x-a(x-1).$
$①当a=4时,求曲线y=f(x)在(1,f(1))处的切线方程;$
$②若当x\in (1,+\infty)时,f(x)\gt 0,求a的取值范围.\Rightarrow 必要性探路$
$例1.②s要证\cfrac{ax^2+x-1}{e^x}+e\ge 0$
$即证:G(a)=H(x)=ax^2+x-1+e^{x+1}\ge 0$
$既可以看作关于x的函数,也可以看作关于a的函数,{\color{Red} 变换主元} $
$\because x^2\ge 0,G(a)为关于a的增函数,变换主元法$
$H(x)_{min}=G(a)_{min}=G(1)=x^2+x-1+{\color{Red }e^{x+1}}\ge x^2+x-1+{\color{Red}x+2}=(x+1)^2\ge 0$
${\color{Red}注意!考试要单独证明e^{x+1}\ge x+2 } $

飘带函数及不等式：

${\color{Red} 对数均值不等式\sqrt{x_1x_2} \lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2} \lt\cfrac{x_1+x_2}{2}$

$f(x)=\frac{1}{2}(x-\frac{1}{x})} $

$g(x)=\frac{2(x-1)}{x+1}$

${\color{Red} \cfrac{1}{2}(x-\cfrac{1}{x} )\le \ln x \le \cfrac{2(x-1)}{x+1} \quad x\in (0,1]} \quad 先证后用$

${\color{Purple} \cfrac{2(x-1)}{x+1}\le \ln x \le \cfrac{1}{2}(x-\cfrac{1}{x} ) \quad x\in [1,+\infty) } \quad 先证后用$
飘带放缩及对数均值不等式

证明：${\color{Red} f(x)=\ln x-\cfrac{2(x-1)}{x+1}}$
${f}' (x)=\cfrac{1}{x}-\cfrac{2(x+1-x+1)}{(x+1)^2}= \cfrac{1}{x}-\cfrac{4}{(x+1)^2}$
$=\cfrac{(x+1)^2-4x}{x(x+1)^2}=\cfrac{(x-1)^2}{x(x+1)^2}\ge 0$
${\color{Violet} \because \quad } f(x)\nearrow ,f(1)=0$
$\Rightarrow {\color{Red} x\in (0,1] } ,\quad f(x)\le 0\quad \ln x\le \cfrac{2(x-1)}{x+1};$
${\color{Green} x\in [1,+\infty)},\quad f(x)\ge 0\quad \ln x\ge \cfrac{2(x-1)}{x+1}$
$再证左边不等式：{\color{Purple} g(x)=\ln x- \cfrac{1}{2}(x-\cfrac{1}{x} )}$
${g}' (x)=\cfrac{1}{x}-\cfrac{1}{2}(1+\cfrac{1}{x^2} )= - \cfrac{1}{2}(1+\cfrac{1}{x^2}-\frac{2}{x})= - \cfrac{1}{2}\cdot \cfrac{(x-1)^2}{x^2}$
${g}' (x)\le 0,\quad g(x)\searrow\qquad g(1)=0$
$\Rightarrow {\color{Red} x\in (0,1]} ,\quad g(x)\ge 0, \quad \ln x \ge \cfrac{1}{2}(x-\cfrac{1}{x} )$
${\color{Green} x\in [1,+\infty)}, \quad g(x)\le 0, \quad \ln x \le \cfrac{1}{2}(x-\cfrac{1}{x} )$

$例1、已知x\gt 0,证明(e^x-1)\ln (x+1)\gt x^2$
$解：先进行放缩e^x\ge 1+x,显然是不够精度的，e^x\ge 1+x+\cfrac{1}{2}x^2$
$得，即证(x+\cfrac{1}{2}x^2)\ln (x+1)\gt x^2$
$即证:(1+\cfrac{1}{2}x)\ln (x+1)\gt x$
$令t=x+1,t\gt 1,x=t-1即证 \cfrac{1}{2}(t+1)\ln t\gt t-1$
$即证:\ln t \gt \cfrac{2(t-1)}{t+1}\quad x\in (1,+\infty)$
${\color{Red} 再让我们来看看飘带不等式与对数不等式的联系： }$
$\sqrt{x_1x_2}\lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2 } \lt \cfrac{x_1+x_2}{2}$
$若要证：\cfrac{x_1-x_2}{\ln x_1-\ln x_2 } \lt \cfrac{x_1+x_2}{2}$
$设x_1\gt x_2,即证\quad\cfrac{x_1-x_2}{x_1+x_2} \lt \cfrac{\ln x_1-\ln x_2 }{2}$
$齐次化上式：得到\quad \cfrac{\cfrac{x_1}{x_2}-1 }{\cfrac{x_1}{x_2}+1 } \lt \cfrac{\ln \cfrac{x_1}{x_2} }{2}$
$令t=\cfrac{x_1}{x_2}\quad t\gt 1换元得,\quad{\color{Red}\cfrac{2(t-1)}{t+1}\lt \ln t }$
${\color{Red} 这便是飘带不等式．}$
$若要证:\sqrt{x_1x_2}\lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2 }$
$设x_1\gt x_2,即证\quad\ln x_1-\ln x_2 \lt \cfrac{x_1-x_2}{\sqrt{x_1x_2}}$
$齐次化上式：得到\ln \cfrac{x_1}{x_2}\lt \cfrac{\cfrac{x_1}{x_2}-1}{\sqrt{\cfrac{x_1}{x_2}}}$
$令t=\sqrt{\cfrac{x_1}{x_2}},\quad t\gt 1换元得，\quad 2\ln t\lt \cfrac{t^2-1}{t}=t-\cfrac{1}{t}$
$即证:\quad {\color{Red}\ln t\lt \cfrac{1}{2}\cdot(t-\cfrac{1}{t}) }$
${\color{Green} 妥妥的飘带不等式．}$

$例2.已知函数f(x)=\ln x -ax^2+(2-a)x$
$(1).求单调性；$
$(2).设f(x)有两个零点,是x_1,x_2,求证x_1+x_2\gt \cfrac{2}{a}$
$(3).设x_0=\cfrac{x_1+x_2}{2},求证:{f}' (x_0)\lt 0$
https://one.free.nf/index.php/archives/200/ $\quad例6$
$f(x)=\ln x-ax^2+(2-a)x$
${f}'(x)=\cfrac{1}{x}-2ax+2-a=\cfrac{-2ax^2+(2-a)x+1}{x}$
$=\cfrac{(2x+1)(-ax+1)}{x},2x+1\gt 0,只需考虑-ax+1即可$
$①a\le 0,{f}' (x)\gt 0,f(x)\nearrow ;$
$②a\gt 0,x\in (0,\cfrac{1}{a}), {f}' (x)\gt 0,f(x)\nearrow ;$
$x\in (\cfrac{1}{a},+\infty), {f}' (x)\lt 0,f(x)\searrow ;$
$f(x)\le f(\cfrac{1}{a})=\ln \cfrac{1}{a}-\cfrac{1}{a}+(2-a)\times\cfrac{1}{a}=\cfrac{1}{a}-\ln a-1$
$设g(a)=\cfrac{1}{a}-\ln a-1,g(a)\searrow 且g(1)=0,所以a\in (0,1)g(a)\gt0,$
$(2).设f(x)有两个零点,是x_1,x_2,求证x_1+x_2\gt \cfrac{2}{a}$
$证：0\lt a \lt 1时有两个零点$
$\ln x_1=ax_1^2-(2-a)x_1\quad ①$
$\ln x_2=ax_2^2-(2-a)x_2\quad ②$
$\Rightarrow ①- ②=\ln\cfrac{x_1}{x_2}=a(x_1^2-x_2^2)-(2-a)(x_1-x2) $
$a(x_1^2-x_2^2+x_1-x2) =\ln\cfrac{x_1}{x_2}+2(x_1-x_2)$
$\cfrac{1}{a} =\cfrac{x_1^2-x_2^2+x_1-x_2}{\ln\cfrac{x_1}{x_2}+2(x_1-x_2)}$
$要证x_1+x_2\gt \cfrac{2}{a}即证x_1+x_2\gt \cfrac{2(x_1^2-x_2^2+x_1-x_2)}{\ln\cfrac{x_1}{x_2}+2(x_1-x_2)} $
$\Leftrightarrow \ln\cfrac{x_1}{x_2}+2(x_1-x_2)\gt \cfrac{2(x_1^2-x_2^2+x_1-x_2)}{x_1+x_2}$
$\Leftrightarrow\ln\cfrac{x_1}{x_2}\gt \cfrac{2(x_1^2-x_2^2+x_1-x_2)}{x_1+x_2}-2(x_1-x_2)=\cfrac{2(x_1-x_2)}{x_1+x_2}$

$例3.已知函数f(x)=\cfrac{\ln x}{x},若f(x)=a有两个不同的零点，试证明：$
$1.\quad 单调性；2.\quad a的取值范围；3.\quad\cfrac{2}{a} \lt x_1+x_2\lt \cfrac{-2\ln a}{a},$
$4.\quad e^2\lt x_1x_2\lt \cfrac{1}{a^2};\quad 5.\quad 2x_1+x_2\lt \cfrac{3}{a} 6.\quad\cfrac{1}{x_1}+\cfrac{1}{x_2}\gt \cfrac{2}{a}$
$7.\quad x_1x_2\gt \cfrac{e}{a}, \quad 8.\quad x_1+x_2\cfrac{3}{a}-e\quad 9.\quad\ln x_1+\ln x_2\gt 1-\ln a或x_1+x_2\gt \cfrac{1-\ln a}{a}$
$10.\quad x_1^2x_2+x_1x_2^2\gt 2 \quad 11.\quad x_1\gt \cfrac{1+\sqrt{1-ax} }{a} \quad 12.\quad x_2\lt \cfrac{1-\sqrt{1-ax} }{a}$

$证：f(x)=a有两个不同的零点\Rightarrow \begin{cases} \quad \ln x_1=ax_1\quad① \\\quad \ln x_2=ax_2\quad②\end{cases}，两式相减最常用！相加何时用到？$
$①-②,\ln x_1-\ln x_2=a(x_1-x_2) \Rightarrow {\color{Red} \cfrac{1}{a}=\cfrac{x_1-x_2}{\ln x_1-\ln x_2} ，对数均值不等式}$
$证3：\quad\cfrac{2}{a} \lt x_1+x_2\lt \cfrac{-2\ln a}{a},和4.\quad e^2\lt x_1x_2\lt \cfrac{1}{a^2};$
$先看4式左右两边求对数，得2\lt \ln x_1+\ln x_2\lt -2\ln a,①+②,得\ln x_1+\ln x_2=a(x_1+x_2)$
${\color{Green}可见3.\quad\cfrac{2}{a} \lt x_1+x_2\lt \cfrac{-2\ln a}{a},4.\quad e^2\lt x_1x_2\lt \cfrac{1}{a^2};式是同一命题 }$
$我们先证3式左边不等式:x_1+x_2\gt \cfrac{2}{a}(消a)\quad \Rightarrow x_1+x_2\gt \cfrac{2}{a}=\cfrac{2(x_1-x_2)}{\ln \cfrac{x_1}{x_2} }$
$令x_1\gt x_2,即证 \ln \cfrac{x_1}{x_2}\gt \cfrac{2(x_1-x_2)}{x_1+x_2}{\quad\color{Red} 交换位置前对数均值，交换后是飘带} $
$设t=\cfrac{x_1}{x_2}\gt 1,即证\ln t\gt \cfrac{2(t-1)}{t+1}$
又是妥妥的飘带不等式;
$再证4式右侧不等式：x_1x_2\lt \cfrac{1}{a^2}$
$x_1x_2\lt \cfrac{1}{a^2},两边开方，得\sqrt{x_1x_2}\lt \cfrac{1}{a} =\cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }$
$\sqrt{x_1x_2}\lt \cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }\Rightarrow \sqrt{x_1x_2}\lt \cfrac{x_1-x_2}{\ln x_1-\ln x_2}$
$①+②,\ln x_1+\ln x_2=a(x_1+x_2) \gt \cfrac{2}{a}\times a=2$
要证6：$\cfrac{1}{x_1}+\cfrac{1}{x_2}\gt \cfrac{2}{a}$
$\cfrac{1}{a}=\cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }$
$即证：\cfrac{1}{x_1}+\cfrac{1}{x_2}\gt 2\times \cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }$
$令x_1\gt x_2\Rightarrow 2\ln \cfrac{x_1}{x_2}\lt (x_1-x_2)(\cfrac{1}{x_1}+\cfrac{1}{x_2}) =\cfrac{x_1}{x_2}-\cfrac{x_2}{x_1}$
$设t=\cfrac{x_1}{x_2},2\ln t\lt t-\cfrac{1}{t}\quad t\gt 1$
这里有错误！
以上难度高二同学掌握足矣！

齐次式+比值换元
$\cfrac{3a-2b}{3a+2b} =\cfrac{3\times \cfrac{a}{b} -2}{3\times \cfrac{a}{b} +2}$
还是天津2010年 高考题目
$f(x)=\cfrac{x}{e^x} ,若x_1\ne x_2,且有f(x_1)=f(x_2)证明：x_1+x_2\gt 2 及 x_1x_2\lt 1$

$\cfrac{3a-2b}{3a+2b} =\cfrac{3\times \cfrac{a}{b} -2}{3\times \cfrac{a}{b} +2}$
$\cfrac{x_1}{e^{x_1}} =\cfrac{x_2}{e^{x_2}}$
$\Rightarrow \cfrac{x_1}{x_2} =\cfrac{e^{x_1}}{e^{x_2}}\Rightarrow {\color{Red} \ln } \cfrac{e^{x_1}}{e^{x_2}}={\color{Red} \ln }\cfrac{x_1}{x_2}\Rightarrow x_1-x_2=\ln \cfrac{x_1}{x_2}$
$\Rightarrow \cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} } ={\color{Red} 1} \quad 1的妙用来了$
$欲证x_1+x_2\gt 2\Leftrightarrow x_1+x_2\gt 2\cdot \cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }$
$设x_1\gt x_2\Rightarrow \ln \cfrac{x_1}{x_2}\gt \cfrac{2(x_1-x_2)}{x_1+x_2}\quad$下一步齐次化
$\Rightarrow \ln \cfrac{x_1}{x_2}\gt \cfrac{2(\cfrac{x_1}{x_2} -1)}{\cfrac{x_1}{x_2} +1}$
设$t=\cfrac{x_1}{x_2}换元得，\ln t\gt \cfrac{2(t-1)}{t+1}$
这是常用的飘带放缩，构成函数求导易证。$t\gt 1$
$g(t)=\ln t-\cfrac{2(t-1)}{t+1}=\ln t-\cfrac{2(t+1-2)}{t+1}=\ln t -2+\cfrac{4}{t+1}$
${\color{Orange}即证t\gt 1时，g(t)_{min}\gt 0}$
${g}'(t)= \cfrac{1}{t} -\cfrac{4}{(t+1)^2} =\cfrac{t^2+2t+1-4t}{t(t+1)^2} \gt 0$
${\color{Orange}g(t)\gt g(t)_{min}=g(1)=0}$

若要证$x_1x_2\lt 1$
$\sqrt{x_1x_2} \lt 1=\cfrac{x_1-x_2}{\ln \cfrac{x_1}{x_2} }\Rightarrow \ln \cfrac{x_1}{x_2} \lt \cfrac{x_1-x_2}{\sqrt{x_1x_2}}\Rightarrow \ln \cfrac{x_1}{x_2}\lt\cfrac{\cfrac{x_1}{x_2}-1 }{\sqrt{\cfrac{x_1}{x_2}} }$
$设t=\sqrt{\cfrac{x_1}{x_2}},\quad t\gt 1\quad \ln t^2\lt \cfrac{t^2-1}{t}=t-\cfrac{1}{t}\Rightarrow \ln t\lt\cfrac{1}{2} (t-\cfrac{1}{t} )$
$这又是飘带不等式，构造h(t)=\cfrac{1}{2} (t-\cfrac{1}{t} )-\ln t\qquad$
${\color{Orange} 即证h(t)_{min}\gt 0}\quad t\gt 1$
${h}' (t)=\cfrac{1}{2}[1+\cfrac{1}{t^2}-\cfrac{2}{t}]=\cfrac{(1+t)^2}{2t^2}\gt 0\Rightarrow{h}(t)\nearrow$
$h(t)\gt {\color{Orange}h(t)_{min}=h(1)=0}$

例：$f(x)=\ln x -ax -1(a\in R),若f(x)+2=0$
有两个不等实根,且$x_2\gt 2x_1$,$求证:x_1x_2^2\gt \cfrac{32}{e^3}，\ln 2=0.693$
$f(x)=\ln x-ax-1\Leftrightarrow f(x)+2=\ln x-ax+1=0\Rightarrow \begin{cases} \quad \ln x_1+1=ax_1\\ \qquad \\ \quad \ln x_2+1=ax_2\end{cases}$
$\Rightarrow \cfrac{ \ln x_1+1}{\ln x_2+1} =\cfrac{x_1}{x_2} \quad 令x_1\lt x_2,\quad t=\cfrac{x_1}{x_2}$

$\Rightarrow x_1=tx_2,\qquad \ln x_1=\ln t+\ln x_2$
$\Rightarrow {\color{Red} t=\cfrac{\ln t+\ln x_2+1}{\ln x_2+1} }=\cfrac{\ln t}{\ln x_2+1} +1$

$$\Rightarrow \begin{cases} \quad \ln x_2=\cfrac{\ln t}{t-1} +1\\ \qquad \\ \quad \ln x_1=\ln t+\ln x_2=\cfrac{t\ln t}{t-1} +1\end{cases}$$
$要证{\color{Green} x_1\cdot x_2^2\gt \cfrac{32}{e^3}} \quad 即证\Leftrightarrow \ln x_1+2\ln x_2\gt 5\ln 2-3$
$即证:\cfrac{2\ln t}{t-1} +\cfrac{t\ln t}{t-1} \gt 5\ln 2$
$令g(t)=\cfrac{2\ln t}{t-1} +\cfrac{t\ln t}{t-1}=\cfrac{(t+2)\ln t}{t-1},$
$即证0\lt t\lt \cfrac{1}{2}时，{\color{Green} g(t)_{min}} \gt 5\ln 2$
${g}' (t)=\cfrac{(\ln t+1+\cfrac{2}{t} )(t-1)-(t+2)\ln t}{(t-1)^2} =\cfrac{t+1-3\ln t-\cfrac{2}{t} }{(t-1)^2}$
$分子太长，单独拎出来，h(t)=t+1-3\ln t-\cfrac{2}{t}$
${h}' (t)=1-\cfrac{3}{t}+\cfrac{2}{t^2}=\cfrac{(t-2)(t-1)}{t^2}\gt 0\quad 0\lt t\lt \cfrac{1}{2}$
$h(\cfrac{1}{2} )=\cfrac{3}{2}-3\ln \cfrac{1}{2}-4= -\cfrac{5}{2}+3\ln 2\lt 0$
$\Rightarrow {g}' (t)\lt 0\Rightarrow g(t)\searrow g(t)\gt g(t)_{min}=g(\cfrac{1}{2})=5\ln 2$