切线放缩辅助分析
设\(f(x)=ax-(a+1)\ln x-\dfrac{1}{x},a>0\)
\((1)\) 讨论\(f(x)\)的单调性
\((2)\) 设\(g(x)=x^2e^{2x}-f(x)\)，若关于\(x\)的不等式\(g(x)\geq ax+(a+3)\ln x+\dfrac{1}{x}+1\)恒成立，求实数\(a\)的取值范围.
解
\((1)\) \(f^{\prime}(x)=a-\dfrac{a+1}{x}+\dfrac{1}{x^2}=\dfrac{ax^2-(a+1)x+1}{x^2}=\dfrac{(ax-1)(x-1)}{x^2}=0\)
即\(x_1=\dfrac{1}{a},x_2=1\)
当\(a>1\)时，\(f(x)\)在\(\left(0,\dfrac{1}{a}\right)\)和\((1,+\infty)\)上单调递增，\(\left(\dfrac{1}{a},1\right)\)单调递减
当\(0<a<1\)时，\(f(x)\)在\((0,1)\)和\(\left(\dfrac{1}{a},+\infty\right)\)上单调递增，\(\left(1,\dfrac{1}{a}\right)\)单调递减
\((2)\) 由题\(g(x)=x^2e^{2x}-ax+(a+1)\ln x+\dfrac{1}{x}\)
得题不等式转化为
\(x^2e^{2x}\geq 2ax+2\ln x+1\)
即\((xe^x)^2\geq \ln\left(xe^{ax}\right)^2+1\)
因\(\ln x\leq x-1\)
当\(a\leq 1\)时，\(\ln\left(xe^{ax}\right)+1\leq\left(xe^{ax}\right)^2-1+1=(xe^{ax})^2\leq(xe^x)^2\)
不等式得证.
\((xe^x)^2-2\ln x-2x\geq 2ax-2x+1\)
即\((xe^x)^2-\ln(xe^x)^2\geq 2ax-2x+1=2x(a-1)+1\)
即\((xe^x)-\ln(xe^x)^2-1\geq 2x(a-1)\)
当\(a>1\)时，记\(xe^x=t\)，左边\(h(t)=t–\ln t^2-1\)
\(h^{\prime}(t)=1-\dfrac{2}{t}\)
则不难得到\(h(t)\)在\((0,2)\)单调递减，\((2,+\infty)\)上单调递增
而\(h(2)=2-\ln4-1<0,t\to+\infty,h(t)\to+\infty\)
从而一定有一点\(t_0,h(_0)=0\)
而\(2x(a-1)>0\)
不合题！
所以\(a\leq 1\)