\(\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right),\ln x>\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\)放缩
已知函数\(f(x)=e^{\frac{1}{x}-a}+\ln x-a\)有两个零点\(x_1,x_2\)
\((1)\) 求\(a\)的取值范围
\((2)\) 证明：\(x_1+x_2>2a\)
\((1)\) \(e^{\frac{1}{x}-a}+\ln x-a=0\)
即\(e^{\frac{1}{x}-a}+\dfrac{1}{x}+\ln x-\dfrac{1}{x}-a=0\)
即\(e^{\frac{1}{x}-a}+\dfrac{1}{x}-a=\dfrac{1}{x}-\ln x=\dfrac{1}{x}+\ln\dfrac{1}{x}=e^{\ln\frac{1}{x}}+\ln\dfrac{1}{x}\)
考虑\(g(x)=e^x+x\)其单调递增
从而\(g\left(\dfrac{1}{x}-a\right)=g\left(\ln\dfrac{1}{x}\right)\)
从而\(\dfrac{1}{x}-a=\ln\dfrac{1}{x}\)有两个零点
即\(a=\dfrac{1}{x}-\ln\dfrac{1}{x}\xlongequal{\frac{1}{x}=t}t-\ln t(t>0)\)有两个零点
记\(h(t)=t-\ln t,h^{\prime}(t)=1-\dfrac{1}{t}\)
从而\(h(t)\)在\((0,1)\)上单调递减，\((1,+\infty)\)上单调递增
从而\(a>1\)有两个零点
\((2)\) 由\((1)\)得\(\begin{cases} a=\dfrac{1}{x_1}-\ln\dfrac{1}{x_1}\\ \\ a=\dfrac{1}{x_2}-\ln\dfrac{1}{x_2} \end{cases}\Rightarrow\begin{cases} ax_1=1+x_1\ln x_1\\ \\ ax_2=1+x_2\ln x_2 \end{cases}\)
不妨设\(0<x_1<1<x_2\)
因\(0<x<1\)时\(\ln x>\dfrac{1}{2}\left(x-\dfrac{1}{x}\right),x>1\)时\(\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\)
从而\(\begin{cases} ax_1=1+x_1\ln x_1>1+\dfrac{x_1}{2}\left(x_1-\dfrac{1}{x_1}\right)\\ \\ ax_2=1+x_2\ln x_2<1+\dfrac{x_2}{2}\left(x_2-\dfrac{1}{x_2}\right) \end{cases}\)
则\(a(x_1-x_2)>\dfrac{(x_1-x_2)(x_2+x_1)}{2}\)
即\(a<\dfrac{x_1+x_2}{2}\)
即\(x_1+x_2>2a\)