切线放缩

指对切线放缩

${\color{Green} e^x必会不等式\Rightarrow \begin{cases} e^x\ge x+1\quad x=0取＝\\ e^x\ge ex\quad x=1取＝\end{cases}}$
${\color{Red} \ln x必会不等式\Rightarrow \begin{cases} 1-\cfrac{1}{x}\le \ln x\le x-1\quad x=1取＝\\ \ln x\le \cfrac{x}{e}\quad x=e取＝\end{cases}}$
${\color{Purple} 常见的三角放缩：} \sin x \lt x \lt \tan x,x\in(0,\cfrac{\pi}{2})$
其他放缩：
$\ln x \lt \sqrt{x} -\frac{1}{\sqrt{x} }(x\gt1)\qquad \ln x \gt \sqrt{x} -\frac{1}{\sqrt{x} }(0\lt x\lt1)$
$\ln x \lt \cfrac{1}{2}(x-\cfrac{1}{x}) (x\gt1)\qquad \ln x \gt \cfrac{1}{2}(x-\cfrac{1}{x}) (0\lt x\lt1)$
$\ln x \gt \cfrac{2(x-1)}{x+1}) (x\gt1)\qquad \ln x \lt \cfrac{2(x-1)}{x+1}) (0\lt x\lt1)$
$\ln x \gt -\cfrac{1}{2}x^2+2x-\cfrac{3}{2} (x\gt1)\quad \ln x \lt -\cfrac{1}{2}x^2+2x-\cfrac{3}{2} (0\lt x\lt1)$
${\color{Red} e^x\ge 1+x+\cfrac{1}{2}x^2 (x\ge 0) } $
例1：$证明不等式e^x-\ln (x+2)\gt 0恒成立$

$例2：x\gt 0時，證明ex^2-x\ln x\lt xe^x+\cfrac{1}{e}$
$ex^2-x\ln x\lt xe^x+\cfrac{1}{e}\Leftrightarrow e^x+\cfrac{1}{ex} \gt ex-\ln x$
$即证 e^x+\cfrac{1}{ex} \ge ex{\color{Green} +\cfrac{1}{ex} } \gt ex{\color{Green} -\ln x} =ex+\ln \cfrac{1}{x}$
$\cfrac{\ln x}{x}\le \cfrac{1}{e}\Rightarrow \cfrac{x}{e}\ge \ln x \Rightarrow {\color{Green} \cfrac{1}{ex} \ge \ln \cfrac{1}{x}}$

$例3：对于\forall x\gt 0,不等式e^x+x^2-(e+1)x+\cfrac{e}{x}\gt 2成立$
$\because e^x\ge ex$
$\Rightarrow {\color{Green} e^x} +x^2-(e+1)x+\cfrac{e}{x}\ge {\color{Green} ex} +x^2-(e+1)x+\cfrac{e}{x}\gt 2$
$x^2-x+\cfrac{e}{x}\gt 2$

$x^2-2x+x+\cfrac{e}{x}=(x-1)^2-1+ x+\cfrac{e}{x}\ge 2\sqrt{e} -1\gt 2$

例4：$e^x+\cfrac{1}{x}\ge 2-\ln x+x^2+(e-2)x$
用${\color{Green}e^x }+\cfrac{1}{x} \ge {\color{Green} ex+(x-1)^2} +\cfrac{1}{x}\ge 2-\ln x+x^2+(e-2)x$

用曲线代替直线放缩：