14. (Alternative Solution)
Let $ c_n = a_n + b_n $
So $ c_2= (a_1 + \frac{1}{a_1})+ (b_1 + \frac{1}{b_1}) $
It's easy to show that $ c_2 \geq 4 $ by AM-GM inequality.
Consider $ (c_{n+1})^2 $ with $ n \geq 2 $
$ \begin{array}{rcl }(c_{n+1})^2 &= & (a_n + b_n)^2 + 2(a_n + b_n)\big (\frac{1}{a_n}+\frac{1}{b_n}\big ) + \big (\frac{1}{a_n}+\frac{1}{b_n}\big )^2 \\ & \geq & (c_n)^2 +2(4)+ \big (\frac{1}{a_n}+\frac{1}{b_n}\big )^2 \\& > & (c_n)^2+ 8 \end{array}$
From above result, we can rewrite inequality in terms of $ c_2$ to be
$ (c_{2+k})^2 >(c_2)^2 +8k \geq 16+8k $
Clearly, if we select $ k \geq \frac{1000^2}{8}-2 = 124,998. $ It follows immediately that $ c_{2+k} > 1000 $
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