已知 $a_1>0$，$b_1>0$，且对任意 $n\in\mathbb N^{\ast}$，有 $a_{n+1}=a_n+\dfrac{1}{b_n}$，$b_{n+1}=b_n+\dfrac{1}{a_n}$，求证：$a_{50}+b_{50}>20$．
解析 根据题意，有\[(a_{n+1}+b_{n+1})^2=(a_n+b_n)^2+2(a_n+b_n)\left(\dfrac{1}{a_n}+\dfrac{1}{b_n}\right)+\left(\dfrac{1}{a_n}+\dfrac{1}{b_n}\right)^2,\]于是\[(a_{n+1}+b_{n+1})^2>(a_n+b_n)^2+8,\]从而\[(a_{50}+b_{50})^2>(a_2+b_2)^2+384,\]又\[a_2+b_2=a_1+\dfrac{1}{b_1}+b_1+\dfrac{1}{a_1}\geqslant 4,\]于是\[(a_{50}+b_{50})^2>400,\]原命题得证．