881

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

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

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