极值点偏移:对数均值不等式
已知\(a\in\mathbb{R}\),函数\(f(x)=\dfrac{a}{x}+\ln x,g(x)=ax-\ln x-2\).若\(f(x_1)=f(x_2)=2(x_1\neq x_2)\)
(1)求出\(a\)的取值范围
(2)证明:\(\dfrac{1}{x_1}+\dfrac{1}{x_2}>\dfrac{2}{a}\)
解
(1)\(f^{\prime}(x)=\dfrac{x-a}{x^2}\)
当\(a\leq 0\)时，\(f(x)\nearrow\)，不合题
当\(a>0\)时，\(f(x)\)在\((0,a)\searrow,(a,+\infty)\nearrow\)
那么要使得\(f(x_1)=f(x_2)=2(x_1\neq x_2)\)
则必有$f(a)=\ln a+1<2 $得\(0<a<e\)
(2)因\(g\left(\dfrac{1}{x}\right)=f(x)-2\)
从而\(f(x_1)=f(x_2)=2(x_1\neq x_2)\)转化为
\(g\left(\dfrac{1}{x_1}\right)=g\left(\dfrac{1}{x_2}\right)\)
记\(\dfrac{1}{x_1}=t_1,\dfrac{1}{x_2}=t_2\),
由(1)知，\(0<x_1<\dfrac{1}{a}<x_2\)
从而\(0<t_2<a<t_2\)
因\(g(t_1)=g(t_2)\)，则\(\begin{cases} ax_1-\ln x_1-2=0\\ ax_2-\ln x_2-2=0 \end{cases}\)
两式相减得\(a(t_1-t_2)=\ln t_1-\ln t_2\)
即\(\dfrac{1}{a}=\dfrac{t_1-t_2}{\ln t_1-\ln t_2}\)
由对数均值不等式
\begin{equation}
\dfrac{a+b}{2}\geq \dfrac{a-b}{\ln a-\ln b}\geq \sqrt{ab}\label{ds}
\end{equation}
知:
\(\dfrac{t_1-t_2}{\ln t_1-\ln t_2}<\dfrac{t_1+t_2}{2}\)
从而\(\dfrac{t_1+t_2}{2}>\dfrac{1}{a}\)
从而\(\dfrac{t_1+t_2}{2}>\dfrac{2}{a}\)
从而\(\dfrac{1}{x_1}+\dfrac{1}{x_2}>\dfrac{2}{a}\)