分析不等式

设函数\(f(x)=\dfrac{1}{2}ax^2+\cos x-1\)

\((1)\) 当\(a\geq 1\)时，证明:\(f(x)\geq 0\)

\((2)\) 证明:\(\dfrac{1}{\tan 1}+\dfrac{1}{2\tan\dfrac{1}{2}}+\dfrac{1}{3\tan\dfrac{1}{3}}+\cdots +\dfrac{1}{n\tan\dfrac{1}{n}}>n-\dfrac{2n}{2n+1}\)
解

\((1)\) \(f^{\prime}(x)=ax-\sin x,f^{\prime\prime}(x)=a-\sin x>0\)

从而\(f^{\prime}(x)\)单调递增，而\(f^{\prime}(0)=0\)，则\(f(x)\)在\((-\infty,0)\)上递减，在\((0,+\infty)\)上递增

进而\(f(x)\geq f(0)=0\)

\((2)\)
分析\(\displaystyle\sum\limits_{i=1}^{n}\dfrac{1}{i\tan \dfrac{1}{i}}=\sum\limits_{i=1}^{n}\dfrac{1}{i}\dfrac{1}{\tan\dfrac{1}{i}}=\sum\limits_{i=1}^{n}\dfrac{1}{i}\dfrac{\cos\dfrac{1}{i}}{\sin\dfrac{1}{i}}\)
由\((1)\)知，取\(a=1\)，则\(\dfrac{1}{2}x^2+\cos x-1\geq 0\)

即\(\cos x\geq 1-\dfrac{1}{2}x^2,x\in(0,1]\)
考虑\(x\dfrac{\cos x}{\sin x}\geq \dfrac{1-\dfrac{1}{2}x^2}{\sin x}\cdot x>1-\dfrac{1}{2}x^2\)

则\(\dfrac{1}{n\tan\dfrac{1}{n}}>1-\dfrac{1}{2n^2}=1-\dfrac{2}{4n^2}>1-\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

取\(n=1,2\cdots n\)，两边取和有

\(\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k\tan\dfrac{1}{k}}>n-\left(1-\dfrac{1}{2n+1}\right)=n-\dfrac{2n}{2n+1}\)