飘带函数及不等式

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！