已知定义在区间 $(-m,m),m>0$，值域为 $\mathbb R$ 的函数 $f(x)$ 满足：$(1)$ 当 $0<x<m$ 时，$f(x)>0$；$(2)$ 对于定义域内的任何实数 $a,b$ 满足：$f(a+b)=\dfrac{f(a)+f(b)}{1-f(a)f(b)}$．若函数 $f(x)$ 存在反函数 $g(x)$，当 $n\in\mathbb N^\ast$ 时，求证：$$g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{n^2+3n+3}\right)<g\left(\dfrac12\right),$$
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【解析】
令 $a=b=0$，可得$$f(0)=\dfrac{2f(0)}{1-f^2(0)},$$解得 $f(0)=0$．当 $x\in(-m,m)$ 时，$-x\in(-m,m)$，且$$0=f(x-x)=\dfrac{f(x)+f(-x)}{1-f(x)f(-x)},$$因此$$f(x)+f(-x)=0,$$即 $f(x)$ 是定义在 $(-m,m)$ 上的奇函数，于是当 $a,b\in(-m,m)$ 时，有$$f(a-b)=\dfrac{f(a)-f(b)}{1+f(a)f(b)},$$下面用数学归纳法证明：$$\mathrm{NE1}:\qquad g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{n^2+3n+3}\right)\leqslant g\left(\dfrac12\right)-g\left(\dfrac1{n+2}\right),$$当 $n=1$ 时，$$f\left(g\left(\dfrac12\right)-g\left(\dfrac13\right)\right)=\dfrac{\dfrac12-\dfrac13}{1+\dfrac12\cdot\dfrac13}=\dfrac17,$$于是$$g\left(\dfrac17\right)=g\left(\dfrac12\right)-g\left(\dfrac13\right),$$此时不等式 $\mathrm{NE1}$ 成立．
假设当 $n=k$ 时，$\mathrm{NE1}$ 成立，即$$g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{k^2+3k+3}\right)\leqslant g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right),$$则当 $n=k+1$ 时$$f\left[g\left(\dfrac1{k+2}\right)-g\left(\dfrac1{k+3}\right)\right]=\dfrac{\dfrac1{k+2}-\dfrac1{k+3}}{1+\dfrac1{k+2}\dfrac1{k+3}}=\dfrac{1}{(k+1)^2+3(k+1)+3},$$于是$$\begin{split} &g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{k^2+3k+3}\right)+g\left(\dfrac1{(k+1)^2+3(k+1)+3}\right)\\
\leqslant&g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right)+g\left(\dfrac1{(k+1)^2+3(k+1)+3}\right)\\
=&g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right)+g\left(\dfrac1{k+2}\right)-g\left(\dfrac1{k+3}\right)\\
=&g\left(\dfrac12\right)-g\left(\dfrac1{k+3}\right).\end{split}$$也满足不等式 $\mathrm{NE1}$．综上，原不等式得证．
假设当 $n=k$ 时，$\mathrm{NE1}$ 成立，即$$g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{k^2+3k+3}\right)\leqslant g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right),$$则当 $n=k+1$ 时$$f\left[g\left(\dfrac1{k+2}\right)-g\left(\dfrac1{k+3}\right)\right]=\dfrac{\dfrac1{k+2}-\dfrac1{k+3}}{1+\dfrac1{k+2}\dfrac1{k+3}}=\dfrac{1}{(k+1)^2+3(k+1)+3},$$于是$$\begin{split} &g\left(\dfrac17\right)+g\left(\dfrac1{13}\right)+\cdots+g\left(\dfrac1{k^2+3k+3}\right)+g\left(\dfrac1{(k+1)^2+3(k+1)+3}\right)\\
\leqslant&g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right)+g\left(\dfrac1{(k+1)^2+3(k+1)+3}\right)\\
=&g\left(\dfrac12\right)-g\left(\dfrac1{k+2}\right)+g\left(\dfrac1{k+2}\right)-g\left(\dfrac1{k+3}\right)\\
=&g\left(\dfrac12\right)-g\left(\dfrac1{k+3}\right).\end{split}$$也满足不等式 $\mathrm{NE1}$．综上，原不等式得证．
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