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已知函数 ${\displaystyle f(x)}$ 满足 ${\displaystyle f(x+1)={\displaystyle \frac{1}{2}+\sqrt{f(x)-f^2(x)}}}$. 求 ${\displaystyle {\displaystyle \sum _{i=1}^{8}f(i)}}$ 的最大值.

首先可以注意到 ${\displaystyle f(x)\ge {\displaystyle \frac{1}{2}}}$. 如果某个 ${\displaystyle x_0}$ 使得 ${\displaystyle f(x_0)={\displaystyle \frac{1}{2}}}$, 则根据递推式立即解得 ${\displaystyle {\displaystyle f(x_0\pm 1)=1,f(x_0\pm 2)=\frac{1}{2}}}$, 也即总有 ${\displaystyle f(x+2)=f(x)}$.

观察条件式，移项后平方得$$\begin{array}{r}{f}^2(x+1)-f(x+1)+\frac{1}{4}=f(x)-f^2(x),\end{array}$$
令${\displaystyle g(x)=f(x)-f(x)}$, 则$$\begin{array}{rl}& g(x+1)+\frac{1}{4}=-g(x),\\ \Rightarrow & g(x+2)=-g(x+1)-\frac{1}{4}=-(-g(x)-\frac{1}{4})-\frac{1}{4}=g(x),\\ \Rightarrow & [f(x+2)-f(x)][f(x+2)+f(x)-1]=0,\end{array}$$
则或者${\displaystyle f(x+2)+f(x)={\displaystyle 1\Rightarrow f(x+2)=f(x)=\frac{1}{2}\Rightarrow f(x+2)=f(x)}}$，或者直接得出${\displaystyle f(x+2)=f(x)}$，这样，$$\begin{array}{rl}\sum _{i=1}^{8}f(i)& =4[f(1)+f(2)]=4[\sqrt{\frac{1}{4}-{(f(1)-\frac{1}{2})}^2}+(f(1)-\frac{1}{2})+1]\\ & \le 4\sqrt{2[\frac{1}{4}-{(f(1)-\frac{1}{2})}^2+{(f(1)-\frac{1}{2})}^2]}+4=4+2\sqrt{2}.\end{array}$$
取等时${\displaystyle f(1)=\cdots =f(8)={\displaystyle \frac{2+\sqrt{2}}{4}}}$.