Sequence Challenge: Prove $a_{50}+b_{50}>20$

[anemone]

The sequence $\{a_n\}$ and $\{b_n\}$ are such that, for every positive integer $n$, $a_n>0,\,b_n>0,\,a_{n+1}=a_n+\dfrac{1}{b_n}$ and $b_{n+1}=b_n+\dfrac{1}{a_n}$. Prove that $a_{50}+b_{50}>20$.

 

[Opalg]

Spoiler: My solution

Let $c_n = a_n+b_n$. Then $$c_{n+1} = a_{n+1} + b_{n+1} = a_n + b_n + \dfrac{1}{b_n}+ \dfrac{1}{a_n} = c_n + \dfrac{c_n}{a_nb_n}.$$ By the AM-GM inequality, $a_nb_n \leqslant \frac14(a_n+b_n)^2 = \frac14c_n^2.$ Therefore $\dfrac1{a_nb_n} \geqslant \dfrac4{c_n^2}$, and it follows that $$c_{n+1} \geqslant c_n + \dfrac4{c_n}.$$ For $x>0$, the function $f(x) = x + \dfrac4x$ has minimum value $4$ (when $x=2$). It follows that $c_2 = f(c_1) \geqslant 4$.

The next step is to prove by induction that $c_n \geqslant 2\sqrt{2n}$ for all $n \geqslant 2$, with strict inequality except possibly when $n=2$. The base case $n=2$ holds, because $c_2 \geqslant 4$. Suppose that the inequality holds for $n$. Since the function $f(x)$ is an increasing function for all $x\geqslant2$, it follows that $f(c_n) \geqslant f(2\sqrt{2n}).$ Therefore $$c_{n+1}^2 = (f(c_n))^2 \geqslant (f(2\sqrt{2n}))^2 = \left(2\sqrt{2n} + \dfrac4{2\sqrt{2n}}\right)^2 = 8n + 8 + \dfrac2n > 8(n+1).$$ Take square roots of each side to get $c_{n+1} > 2\sqrt{2(n+1)}$, which completes the inductive step.

Finally, take $n=50$ to see that $a_{50}+b_{50} = c_{50} > 2\sqrt{100} = 20$.