来个简单的
已知函数\(f(x)=2a\ln x-x+\dfrac{1}{x}\)
\((1)\) 若\(\forall x\in [1,+\infty),f(x)\leq 0\)，求\(a\)的取值范围.
\((2)\)证明：\(\forall a\in (1,+\infty),\forall x\in(1,+\infty),f(x)>-(x-1)^2\)
\((1)\) \(2a\ln x-x+\dfrac{1}{x}\leq 0\)即\(2a\leq \dfrac{x^2-1}{x\ln x}\)
记\(g(x)=\dfrac{x^2-1}{x\ln x},g^{\prime}(x)=\dfrac{2x^2\ln x-(x^2-1)(\ln x+1)}{(x\ln x)^2}=\dfrac{x^2\ln x-x^2+\ln x+1}{(x\ln x)^2}\)
记\(h(x)=x^2\ln x-x^2+\ln x+1,h^{\prime}(x)=2x\ln x-x+\dfrac{1}{x}\)
因\(x\geq 1,\ln x\leq \dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\)
所以\(\ln x\leq x-\dfrac{1}{x}-x+\dfrac{1}{x}=0\)
从而\(g(x)\)单调递减
且\(\lim\limits_{x\to 1}g(x)=2\)
进而\(2a\leq2\)即\(a\leq 1\)
\((2)\) 要证：\(a>1\)，\(2a\ln x-x+\dfrac{1}{x}>-(x-1)^2\)
成立
即\(2\ln x-x+\dfrac{1}{x}>-(x-1)^2\)
即证：\(2\ln x-x+\dfrac{1}{x}+(x-1)^2>0\)
记\(G(x)=2\ln x-x+\dfrac{1}{x}+(x-1)^2\)
\(G^{\prime}(x)=(x-1)\cdot\dfrac{2x^2-x+1}{x^2}>0\)
从而\(G(x)\)单调递增，进而\(G(x)>G(1)=0\)
不等式得证\(\square\).