给定正实数$x_1,y_1,z_1$，定义数列$\{x_n\},\{y_n\},\{z_n\}$如下：$$x_{n+1}=y_n+\dfrac 1{z_n},y_{n+1}=z_n+\dfrac{1}{x_n},z_{n+1}=x_n+\dfrac 1{y_n},$$求证：$x_{200},y_{200},z_{200}$中至少有一个数大于$20$．
分析与证明 将三式相加，有$$x_{n+1}+y_{n+1}+z_{n+1}=\left(x_n+y_n+z_n\right)+\left(\dfrac 1{x_n}+\dfrac 1{y_n}+\dfrac 1{z_n}\right),$$于是\[\begin{split} &\qquad \left(x_{n+1}+y_{n+1}+z_{n+1}\right)^2-\left(x_n+y_n+z_n\right)^2\\ &= 2\left(x_n+y_n+z_n\right)\left(\dfrac 1{x_n}+\dfrac 1{y_n}+\dfrac 1{z_n}\right)+\left(\dfrac 1{x_n}+\dfrac 1{y_n}+\dfrac 1{z_n}\right)^2.\end{split} \]由柯西不等式可得$$\left(x_n+y_n+z_n\right)\left(\dfrac 1{x_n}+\dfrac 1{y_n}+\dfrac 1{z_n}\right)\geqslant 9,$$于是就有$$\left(x_{n+1}+y_{n+1}+z_{n+1}\right)^2-\left(x_n+y_n+z_n\right)^2>18,$$分别取$n$为$1,2,\cdots ,199$，累加可得$$\left(x_{200}+y_{200}+z_{200}\right)^2-\left(x_1+y_1+z_1\right)^2>18(200-1),$$因此$$x_{200}+y_{200}+z_{200}>\sqrt{18\cdot 199},$$而$\sqrt{18\cdot 199}<60$，放过头了．
略加调整，由于$$x_2+y_2+z_2=x_1+\dfrac {1}{x_1}+y_2+\dfrac{1}{y_1}+z_1+\dfrac{1}{z_1}\geqslant 6,$$于是$$\left(x_{200}+y_{200}+z_{200}\right)^2-\left(x_2+y_2+z_2\right)^2>18\cdot 198,$$从而可得$$x_{200}+y_{200}+z_{200}>\sqrt{18\cdot 198+36}=60,$$于是命题得证．