指对分离：\(x\ln x,xe^x\)，下界大于上界
已知函数\(f(x)=\dfrac{ae^{x-1}}{x}+e(\ln x-x),a\in\mathbb{R}\)
\((1)\)若\(f(x)\)在\((1.+\infty)\)上单调递增，求\(a\)的取值范围
\((2)\)当\(a\geq \dfrac{5}{2}\)时，证明：\(f(x)+(e-1)x>e^{x-1}(1-\ln x)+e\ln x\)
解
\((1)\) \(f^{\prime}(x)=\dfrac{ae^{x-1}(x-1)}{x^2}+e\left(\dfrac{1}{x}-1\right)=\dfrac{(x-1)(ae^{x-1}-ex)}{x^2}\)
即\(ae^{x-1}-ex\geq 0\)
即\(a\geq (xe^{-x})_{\max}=\dfrac{1}{e}\)
\((2)\) 原不等式为\(\dfrac{ae^{x-1}}{x}-x-e^{x-1}(1-\ln x)>0\)
即\(ae^{x-1}-x^2-xe^{x-1}(1-\ln x)>0\)
即证:\(\dfrac{5}{2}e^{x-1}-x^2-xe^{x-1}(1-\ln x)>0\)
即证：\(\dfrac{5}{2}-x^2e^{1-x}-x(1-\ln x)>0\)
即证：\(\dfrac{5}{2}-x^2e^{1-x}>x(1-\ln x)\)
记\(\varphi(x)=\dfrac{5}{2}-x^2e^{1-x}\)，记\(\gamma(x)=x(1-\ln x)\)
\(\varphi^{\prime}(x)=-e^{1-x}(2x-x^2),\gamma^{\prime}(x)=-\ln x\)
从而\(\varphi(x)\)在\((0,2)\)上单调递减，\((2,+\infty)\)单调递增
\(\gamma(x)\)在\((0,1)\)上单调递增，\((1,+\infty)\)上单调递减
从而\(\varphi(x)\geq \varphi(2)=\dfrac{5}{2}-\dfrac{4}{e},\gamma(x)\leq \gamma(1)=1\)
从而\(\gamma(x)\leq 1<\dfrac{5}{2}-\dfrac{4}{e}\leq \varphi(x)\)
不等式得证！\(\square\)
标签：ae,导数,ln,dfrac,每日,varphi,xe,gamma From： https://www.cnblogs.com/manxinwu/p/17904479.html