重要放缩与观察配凑数列
函数\(f(x)=a\ln x+\dfrac{1}{2}x^2-(a+1)x+\dfrac{3}{2}(a>0)\)
\((1)\)求函数单调区间
\((2)\)当\(a=1\)时，\(f(x_1)+f(x_2)=0\)证明：\(x_1+x_2\geq 2\)
\((3)\)求证：\(\forall n\in\mathbb{N}^{*}\)都有\(2\ln(n+1)+\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{i-1}{i}\right)^2>n\)
解
\((1)\) \(f^{\prime}(x)=\dfrac{a}{x}+x-a-1=\dfrac{x^2-(a+1)x+a}{x}=\dfrac{(x-a)(x-1)}{x}=0\)
得\(x_1=1,x_2=a\)
当\(a=1\)时，\(f^{\prime}(x)\geq 0\)得\(f(x)\)单调递增
当\(a>1\)时，\(f(x)\)在\((0,1)\)和\((a,+\infty)\)上单调递增在\((1,a)\)上单调递减
当\(0<a<1\)时，\(f(x)\)在\((0,a)\)和\((1,+\infty)\)上单调递增在\((a,1)\)上单调递减
\((2)\) \(a=1,f(x)=\ln x+\dfrac{1}{2}x^2-2x+\dfrac{3}{2}\)
不难发现\(f(1)=0\)，从而要使得\(f(x_1)+f(x_2)=0\)则要有\(0<x_1<1<x_2\)
则要证:\(x_1+x_2\geq 2\)
即证:\(x_2\geq2-x_1\)
即证:\(f(x_2)\geq f(2-x_1)\)
即证\(f(2-x_1)-f(x_2)=f(2-x_1)+f(x_1)\leq 0\)
\(f(2-x_1)+f(x_1)=\ln(2-x_1)+\dfrac{(2-x_1)^2}{2}-2(2-x_1)+\ln x_1+\dfrac{x_1^2}{2}-2x_1+3\)
\(=\ln x_1(2-x_1)+x_1^2-2x_1+1\leq 2x_1-x_1^2-1+x_1^2-2x_1+1=0\)
得证！
\((3)\) 取\(a=1,\)则由\(f(x)\)单调递增，则对任意的\(x\in[1,+\infty)\)有\(f(x)>f(1)=0\)
即\(\ln x+\dfrac{1}{2}x^2-2x+\dfrac{3}{2}>0\)
即\(2\ln x+x^2-4x+3>0\)
即\(2\ln x+(x-2)^2>1\)
用\(\dfrac{i+1}{i}\)来代换\(x,i=1,2,3\cdots n\)有
\(2\ln\dfrac{i+1}{i}+\left(\dfrac{1-i}{i}\right)^2>1\)
累加\(n\)项便有\(2\ln(n+1)+\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{i-1}{i}\right)^2>n\).
不等式得证!\(\square\)