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已知函数 ${\displaystyle f(x)={\displaystyle 2m\ln x-x+\frac{1}{x}(m>0)}}$.

1. 讨论 ${\displaystyle f(x)}$ 的单调性;
2. 证明: ${\displaystyle {\displaystyle (1+\frac{1}{2^2})(1+\frac{1}{3^2})(1+\frac{1}{4^2})\cdots (1+\frac{1}{n^2})<{\mathrm{e}}^{\frac{2}{3}}(n\in {\mathbb{N}}^{\ast },n\ge 2)}}$;
3. 若函数 ${\displaystyle {\displaystyle g(x)=m^2{\ln }^{2}x-x-\frac{1}{x}+2}}$ 有三个不同的零点，求 ${\displaystyle m}$ 的取值范围.

1. 解 ${\displaystyle f^{\prime }(x)={\displaystyle -\frac{x^2-2mx+1}{x^2},x>0}}$. 当 ${\displaystyle 0<m\le 1}$ 时，$$f^{\prime }(x)\le -\frac{x^2-2x+1}{x^2}\le 0,$$ 故 ${\displaystyle f(x)}$ 在 ${\displaystyle (0,+\mathrm{\infty })}$ 上减. 当 ${\displaystyle m>1}$ 时，${\displaystyle f^{\prime }(x)}$ 有零点 ${\displaystyle {\displaystyle x_1=m-\sqrt{m^2-1}<1,x_2=m+\sqrt{m^2-1}>1}}$, 则 ${\displaystyle f(x)}$ 在 ${\displaystyle (0,x_1),(x_2,+\mathrm{\infty })}$ 上减, 在 ${\displaystyle (x_1,x_2)}$ 上增.

2. 证明 当 ${\displaystyle m=1}$，由1，${\displaystyle f^{\prime }(x)\le 0}$, 则 ${\displaystyle x>1}$ 时 ${\displaystyle f(x)>f(1)=0\Rightarrow {\displaystyle \ln x<\frac{1}{2}(x-\frac{1}{x})}}$. 从而 $$\begin{array}{rl}\sum _{i=2}^n\ln (1+\frac{1}{i^2})<& \frac{1}{2}\sum _{i=2}^n(\frac{i^2+1}{i^2}-\frac{i^2}{i^2+1})=\frac{1}{2}\sum _{i=2}^n(\frac{1}{i^2}+\frac{1}{i^2+1})\\ <& \sum _{i=2}^n\frac{1}{i^2}<\frac{1}{4}+\sum _{i=3}^n\frac{1}{i^2-1}=\frac{1}{4}+\frac{1}{2}\sum _{i=3}^n(\frac{1}{i-1}-\frac{1}{i+1})\\ =& \frac{1}{4}+\frac{1}{2}(\frac{1}{3}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1})<\frac{1}{4}+\frac{1}{2}(\frac{1}{3}+\frac{1}{2})=\frac{2}{3},\end{array}$$
   两边同时作为 ${\displaystyle \mathrm{e}}$ 的指数，即可证明原不等式.

3. 解 记 ${\displaystyle \sqrt{x}=t}$, 则$$\begin{array}{r}g(x)=g(t^2)=4m^2{\ln }^{2}t-{(t-\frac{1}{t})}^2=(2m\ln t+t-\frac{1}{t})(2m\ln t-t+\frac{1}{t}),\end{array}$$
   记${\displaystyle h(t)=2m\ln t+t-{\displaystyle \frac{1}{t}}}$, 则${\displaystyle {\displaystyle h^{\prime }(t)=\frac{2m}{t}+1+\frac{1}{t^2}>0}}$. 又${\displaystyle h(1)=0}$, 故${\displaystyle h(t)}$有唯一零点${\displaystyle 1}$，它也是${\displaystyle f(t)}$的零点. 这样${\displaystyle g(t^2)=f(t)h(t)=0⟺f(t)=0}$.
   当${\displaystyle 0<m\le 1}$, 由1得${\displaystyle f(t)}$单减，不会有三个零点; 当${\displaystyle m>1}$, 首先依据${\displaystyle \ln x}$在${\displaystyle 4m}$处的切线放缩得到${\displaystyle \ln x\le {\displaystyle \frac{1}{4m}x+\ln 4m-1}}$, 有$$\begin{array}{r}f(t)>-2m\ln \frac{1}{t}-1+\frac{1}{t}\ge -2m(\frac{1}{4mt}+\ln 4m-1)-1+\frac{1}{t}=\frac{1}{2t}-2m(\ln 4m-1)-1,\end{array}$$
   取 ${\displaystyle x_3=min{\displaystyle \{x_1,\frac{1}{2(2m(\ln 4m-1)-1)}\}<1}}$ 得 ${\displaystyle f(x_3)>0}$, 而 ${\displaystyle f(x_1)<f(1)=0}$, 故 ${\displaystyle f(x)}$ 在 ${\displaystyle (x_3,x_1)}$ 上存在一零点.
   其次，$$\begin{array}{r}f(t)<2m(\frac{1}{4m}t+\ln 4m-1)-t+1=-\frac{t}{2}+2m(\ln 4m-1)+1,\end{array}$$
   取 ${\displaystyle x_4={\displaystyle max\{x_2,2(2m(\ln 4m-1)+1)\}>1}}$, 有${\displaystyle f(x_4)<0}$, 而${\displaystyle f(x_2)>f(1)>0}$, 故${\displaystyle f(x)}$在${\displaystyle (x_2,x_4)}$上存在一零点，再结合${\displaystyle 1}$，此时${\displaystyle f(x)}$确实存在三个零点.

   综上，${\displaystyle m>1}$.