2015年全国高中数学联赛浙江省预赛

略

因为$$a_{n+1}^2+b_{n+1}^2=a_n^2+b_n^2+\dfrac{1}{a_n^2}+\dfrac{1}{b_n^2}+2\left(\dfrac{a_n}{b_n}+\dfrac{b_n}{a_n}\right),$$所以\[\begin{split}a_{50}^2+b_{50}^2&=a_1^2+b_1^2+\sum\limits_{i=1}^{49}{\left(\dfrac{1}{a_i^2}+\dfrac{1}{b_i^2}\right)}+2\sum\limits_{i=1}^{49}{\left(\dfrac{a_i}{b_i}+\dfrac{b_i}{a_i}\right)}\\&>a_1^2+b_1^2+\dfrac{1}{a_1^2}+\dfrac{1}{b_1^2}+2\cdot2\cdot49\\&\geqslant200.\end{split}\]又因为$$a_{n+1}b_{n+1}=a_nb_n+\dfrac{1}{a_nb_n}+2,$$所以$$\begin{split}a_{50}b_{50}&=a_1b_1+\sum\limits_{i=1}^{49}\dfrac{1}{a_ib_i}+2\cdot 49\\&>98+a_1b_1+\dfrac{1}{a_1b_1}\\&\geqslant100,\end{split}$$故$$(a_{50}+b_{50})^2=a_{50}^2+b_{50}^2+2a_{50}b_{50}>400,$$因此 $a_{50}+b_{50}>20$．