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已知 ${\displaystyle A(-1,0)}$, ${\displaystyle B(1,0)}$, ${\displaystyle \mathrm{\Delta }MAB}$ 中, ${\displaystyle M}$ 为动点, 记 ${\displaystyle \mathrm{\angle }MAB=\alpha }$, ${\displaystyle \mathrm{\angle }MBA=\beta }$, ${\displaystyle M}$ 满足条件: ${\displaystyle \cos \alpha +\cos \beta =k\sin (\alpha +\beta )}$ (${\displaystyle k}$ 为常数, ${\displaystyle k>1}$). 记动点 ${\displaystyle M}$ 轨迹上的点到原点的距离最小值为 ${\displaystyle \sqrt{f(k)}}$.

1. 求 ${\displaystyle M}$ 点的轨迹方程;
2. 求 ${\displaystyle f(k)}$ 的解析式;
3. 求证 ${\displaystyle f(2)+f(3)+\cdots +f(n)<\frac{3}{4}}$.

1. 解 设 ${\displaystyle M(x,y)}$, 则 ${\displaystyle \cos \alpha =\frac{x+1}{\sqrt{(x+1{)}^2+y^2}}}$, ${\displaystyle \cos \beta =\frac{1-x}{\sqrt{(1-x{)}^2+y^2}}}$. 再由正弦定理 $$\sin (\alpha +\beta )=\sin \beta \frac{2}{AM}=\frac{2y}{\sqrt{(x+1{)}^2+y^2}\sqrt{(x-1{)}^2+y^2}},$$ 于是条件式变为 $$\begin{array}{rl}& \frac{x+1}{\sqrt{(x+1{)}^2+y^2}}+\frac{1-x}{\sqrt{(1-x{)}^2+y^2}}=\frac{2ky}{\sqrt{[(x+1{)}^2+y^2][(x-1{)}^2+y^2]}}\\ \Rightarrow & (x+1)\sqrt{(x-1{)}^2+y^2}+(1-x)\sqrt{(x+1{)}^2+y^2}=2ky\\ \Rightarrow & (x+1)\sqrt{(x-1{)}^2+y^2}=2ky+(x-1)\sqrt{(x+1{)}^2+y^2}\\ \Rightarrow & (x+1{)}^2{y}^2=4k^2{y}^2+(x-1{)}^2{y}^2+4ky(x-1)\sqrt{(x+1{)}^2+y^2}\\ \Rightarrow & (x-k^2)y=k(x-1)\sqrt{(x+1{)}^2+y^2}\\ \Rightarrow & (x^2+k^4)y^2=k^2(x^2-1{)}^2+k^2(x^2+1)y^2,\end{array}$$ 于是 $$y^2=\frac{k^2(x^2-1{)}^2}{(k^2-1)(k^2-x^2)}.$$
2. 解 记 ${\displaystyle t=k^2-x^2\in (0,k^2]}$. 则 $$\begin{array}{rl}{x}^2+y^2& =x^2+\frac{k^2(x^2-1{)}^2}{(k^2-1)t}=k^2-t+\frac{k^2[(k^2-1)-t{]}^2}{(k^2-1)t}\\ & =\frac{1}{k^2-1}t+\frac{k^2(k^2-1)}{t}-k^2.\end{array}$$ 下面需要判断极值点 ${\displaystyle k(k^2-1)}$ 和 ${\displaystyle t}$ 的上界 ${\displaystyle k^2}$ 的大小关系, 分界点位 ${\displaystyle k=\frac{1+\sqrt{5}}{2}}$, 于是 $$f(k)=\{\begin{array}{rl}& 2k-k^2,k\in (1,\frac{1+\sqrt{5}}{2}],\\ & \frac{1}{k^2-1},k\in (1+\frac{\sqrt{5}}{2},+\mathrm{\infty }).\end{array}$$
3. 证明 由 2 得 $$\begin{array}{rl}f(2)+\cdots +f(n)& =\sum _{i=2}^n\frac{1}{i^2-1}=\frac{1}{2}\sum _{i=2}^n(\frac{1}{i-1}-\frac{1}{i+1})\\ & =\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1})<\frac{3}{4}.\end{array}$$