不同角度解决双变量问题
已知函数\(f(x)=x\ln x-\dfrac{1}{2}ax^2-x(a\in\mathbb{R})\)
\((1)\) 若函数\(f(x)\)在\(\left[\dfrac{1}{e},+\infty\right)\)上为增函数，求实数\(a\)的最大值；
\((2)\) 若\(f(x)\)有两个极值点\(x_1,x_2(x_1<x_2)\)，且不等式\(\dfrac{x_2^m}{e}>\dfrac{e^m}{x_1}\)恒成立，求正数\(m\)的取值范围.
解
\((1)\) \(f^{\prime}(x)=\ln x-ax\geq 0\)恒成立
即\(a\leq \left(\dfrac{\ln x}{x}\right)_{\max}=\dfrac{1}{e}\)
\((2)\) 由题\(\ln x_1-ax_1=\ln x_2-ax_2=0\)
即\(\begin{cases} \dfrac{\ln x_1}{x_1}=a\\ \\ \dfrac{\ln x_2}{x_2}=a \end{cases}\)得\(a=\dfrac{\ln x_1+\ln x_2}{x_1+x_2}=\dfrac{\ln x_1-\ln x_2}{x_1-x_2}\)，并且不难知道\(0<x_1<e<x_2\)
原不等式\(\dfrac{x_2^m}{e}>\dfrac{e^m}{x_1}\)即\(\left(\dfrac{x_2}{e}\right)^m>\dfrac{e}{x_1}\)，两边取对数有\(m\ln\dfrac{x_2}{e}>\ln\dfrac{e}{x_1}\)
即\(-m(1-\ln x_2)>1-\ln x_1\) ，记\(\dfrac{x_1}{x_2}=t\)
即\(-m<\dfrac{1-\ln x_1}{1-\ln x_2}=\dfrac{1-ax_1}{1-ax_2}=\dfrac{1-x_1\cdot\dfrac{\ln x_1-\ln x_2}{x_1-x_2}}{1-x_2\cdot \dfrac{\ln x_1-\ln x_2}{x_1-x_2}}=\dfrac{1-t\cdot\dfrac{\ln t}{t-1}}{1-\dfrac{\ln t}{t-1}}=\dfrac{t-1-t\ln t}{t-1-\ln t}\)
记\(g(t)=\dfrac{t-1-t\ln t}{t-1-\ln t}\)，则\(g^{\prime}(t)=\dfrac{\ln^2t-\dfrac{2}{t}+2}{(t-1-\ln t)^2}\)
因\(\ln^2t-\dfrac{2}{t}+2\)单调递增，所以\(\ln^2t-\dfrac{2}{t}+2<0-2+2=0\)
从而\(g(t)\)单调递减
连续两次利用洛必达法则\(\lim\limits_{t\to 1}g(t)=-1\)
从而\(-m\leq -1\)即\(m\geq 1\).
本题根据思路不同还能避免使用洛必达法则
记\(\dfrac{x_1}{x_2}=t\)对原不等式直接取自然对数有
即
即\(\ln t-\dfrac{(m+1)(t-1)}{t+m}<0\)
记\(\varphi(t)=\ln t-\dfrac{(m+1)(t-1)}{t+m}(t\in (0,1)),\varphi(1)=0\)
\(\varphi^{\prime}(t)=\dfrac{(t-1)(t-m^2)}{t(t+m)^2}\)
若\(m^2\geq 1\),\(\varphi^{\prime}(t)>0\)从而\(\varphi(t)\)在\(t\in(0,1)\)上单调递增，合题
若\(m^2<1\)时，\(\varphi(t)\)在\((0,m^2)\)递增，\((m^2,1)\)递减
而在\(x \in(m^2,1)\)上\(\varphi(t)\)单调递减，\(\varphi(1)=0\)
从而在\(x\in(m^2,1)\)上\(\varphi(t)>0\)不合题！
综上\(m^2\geq 1\)成立，又因\(m\)取正数.
所以\(m\geq 1.\)