已知正数数列$\{x_n\}$与$\{y_n\}$对一切正整数$n$，均满足条件$$\begin{split}x_{n+2}&=x_n+x_{n+1}^2,\\y_{n+2}&=y_n^2+y_{n+1}.\end{split}$$如果$x_1,x_2,y_1,y_2$均大于$1$，试判断当$n$足够大时$x_n$与$y_n$的大小关系，并证明．
解 易知，两个数列都从第$2$项起递增，且容易证明$x_3,y_3>2$，进而对于任意$n>3$，均有$x_n,y_n>3$．
注意到对$n>1$，有$$x_{n+2}>x_{n+1}^2>x_n^4,$$而对于$n>3$，有$$y_{n+2}=y_n^2+y_{n+1}=y_n^2+y_n+y_{n-1}^2<3y_n^2<y_n^3,$$这样就有$$\dfrac{\ln x_{n+2}}{\ln y_{n+2}}>\dfrac 43\cdot \dfrac{\ln x_n}{\ln y_n},$$从而$$\dfrac{\ln x_{2k}}{\ln y_{2k}}>\left(\dfrac 43\right)^{k-1}\dfrac{\ln x_2}{\ln y_2},$$且$$\dfrac{\ln x_{2k+1}}{\ln y_{2k+1}}>\left(\dfrac 43\right)^{k}\dfrac{\ln x_1}{\ln y_1},$$因此当$n$足够大时，有$x_n>y_n$．