极值点偏移-进阶

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$x_2\gt 2x_1$,$求证:x_1x_2^2\gt \cfrac{32}{e^3}，\ln 2=0.693$
$f(x)=\ln x-ax-1\Leftrightarrow f(x)+2=\ln x-ax+1=0\Rightarrow \begin{cases} \quad \ln x_1+1=ax_1\\ \qquad \\ \quad \ln x_2+1=ax_2\end{cases}$
$\Rightarrow \cfrac{ \ln x_1+1}{\ln x_2+1} =\cfrac{x_1}{x_2} \quad 令x_1\lt x_2,\quad t=\cfrac{x_1}{x_2}$

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

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