每日导数17

很难的放缩：对数均值不等式

已知函数\(f(x)=-2x-2\sin x+2m\ln x,m>0\)若存在\(f(x_1)=f(x_2)(x_1\neq x_2)\)
\((1)\)判断\(2(x-\sin x)\)的单调性

\((2)\)证明:\(x_1+x_2>1+\ln m\)
解
\((1)\)令\(g(x)=2(x-\sin x)\),\(g^{\prime}(x)=2-2\cos x>0\)
从而其单调递增

\((2)\) \(f(x_1)=f(x_2)\)即\(-x_1-\sin x_1+m\ln x_1=-x_2-\sin x_2+m\ln x_2\)

即\(x_1-x_2=m\ln\dfrac{x_1}{x_2}+\sin x_2-\sin x_1\)

即\(\sin x_1-\sin x_2=m(\ln x_1-\ln x_2)-(x_1-x_2)\)

即\(\dfrac{\sin x_1-\sin x_2}{x_1-x_2}=\dfrac{m(\ln x_1-\ln x_2)}{x_1-x_2}-1\)

不防设\(x_1<x_2\)，则由(1)得\(x_1-\sin x_1<x_2-\sin x_2\)

从而有\(\dfrac{\sin x_1-\sin x_2}{x_1-x_2}<1\)

则原不等式又能改为
\(\dfrac{m(\ln x_1-\ln x_2)}{x_1-x_2}-1<1\)

即证：
\(\dfrac{\ln x_1-\ln x_2}{x_1-x_2}<\dfrac{2}{m}\)

由对数均值不等式\(\dfrac{a+b}{2}\geq \dfrac{a-b}{\ln a-\ln b}\geq \sqrt{ab}\)

我们有\(\dfrac{\ln a-\ln b}{a-b}\geq \dfrac{2}{a+b}\)

即有\(\dfrac{2}{x_1+x_2}\leq \dfrac{\ln x_1-\ln x_2}{x_1-x_2}<\dfrac{2}{m}\)

即有\(x_1+x_2>m\)

即证：\(m>1+\ln m\)

因\(\ln x<x-1\)

则原不等式得证！