We will prove that $a_n^2+b_n^2+c_n^2 > 6n+39$ Use induction: $a_{n+1}^2+b_{n+1}^2+c_{n+1}^2 > a_n^2+b_n^2+c_n^2+2(\frac{a_n}{b_n}+\frac{b_n}{c_n}+\frac{c_n}{a_n})> 6n+39+6=6n+45$

David-Vieta wrote: We will prove that $a_n^2+b_n^2+c_n^2 > 6n+39$ Use induction: $a_{n+1}^2+b_{n+1}^2+c_{n+1}^2 > a_n^2+b_n^2+c_n^2+2(\frac{a_n}{b_n}+\frac{b_n}{c_n}+\frac{c_n}{a_n})> 6n+39+6=6n+45$ making a problem that looks hard on first sight, look like child's play. Nicely done!

Even simpler: If $m_n=\max(a_n,b_n,c_n)$, then $m_{n+1} \ge m_n+\frac{1}{m_n}$ and hence $m_{n+1}^2 \ge m_n^2+2$ so that $m_n^2 \ge 2n+23$ and hence $m_n \ge \sqrt{2n+23}$ which is quite a bit better.