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！