放缩有度

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．