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设函数 ${\displaystyle f(x)={\displaystyle \frac{1}{2}a{x}^2+\cos x-1}}$.

1. 当 ${\displaystyle a\ge 1}$ 时, 证明: ${\displaystyle f(x)\ge 0}$;
2. 证明: ${\displaystyle {\displaystyle \frac{1}{\tan 1}+\frac{1}{2\tan \frac{1}{2}}+\frac{1}{3\tan \frac{1}{3}}+\cdots +\frac{1}{n\tan \frac{1}{n}}>n-\frac{2n}{2n+1}}}$.

1. 注意到 ${\displaystyle f(x)}$ 是偶函数. 当 ${\displaystyle x\ge 0}$，构造 ${\displaystyle g(x)=x-\sin x}$，则 ${\displaystyle g^{\prime }(x)=1-\cos x\ge 0\Rightarrow g(x)\ge g(0)=0}$. 同理 ${\displaystyle x+\sin x\ge 0}$. 这样 ${\displaystyle x\ge 0}$ 时$$\begin{array}{r}f(x)\ge \frac{1}{2}{x}^2-2{\sin }^2\frac{x}{2}=\frac{1}{2}(\frac{x}{2}-\sin \frac{x}{2})(\frac{x}{2}+\sin \frac{x}{2})\ge 0,\end{array}$$
   再结合偶函数，${\displaystyle f(x)\ge 0}$ 得证.
2. 依然抓住 ${\displaystyle x\ge \sin x,x\ge 0}$，有$$\begin{array}{rl}\frac{1}{i\tan \frac{1}{i}}& =\frac{\cos \frac{1}{i}}{i\sin \frac{1}{i}}=\frac{1-2{\sin }^2\frac{1}{2i}}{i\sin \frac{1}{i}}\ge \frac{1-2{\left(\frac{1}{2i}\right)}^2}{i\left(\frac{1}{i}\right)}\\ & =1-\frac{2}{4i^2}>1-\frac{2}{4i^2-1}=1-(\frac{1}{2i-1}-\frac{1}{2i+1}),\end{array}$$
   从而$$\begin{array}{r}\sum _{i=1}^n\frac{1}{i\tan \frac{1}{i}}>\sum _{i=1}^n(1-\frac{1}{2i-1}+\frac{1}{2i+1})=n-1+\frac{1}{2n+1}=n-\frac{2n}{2n+1}.\end{array}$$