I posted this problem here, but I did not know at the time that it's from a competition.

Let $b_n := a_n/4$. Then we have: \[b_{n+2}-1 = \frac{1}{2} \left( (\sqrt{b_n}-1) + (\sqrt{b_{n+1}} -1) \right) \] Using $\sqrt{b_n}-1 = (b_n-1)/(\sqrt{b_n}+1)$, we get: \[ |b_{n+2}-1| \le \max (|b_n-1|, |b_{n+1}-1| ) \,\,\,\,\,\, \text{(1)} \] \[ |b_{n+2}-1| \le \frac{\max (|b_n-1|, |b_{n+1}-1| )}{ \min (1+\sqrt{b_n},1+\sqrt{b_{n+1}})} \,\,\,\,\,\, \text{(2)} \] $\text{(1)}$ implies $|b_{n}-1| \le \max (|b_0-1|, |b_1-1| )$, therefore $\min (1+\sqrt{b_n},1+\sqrt{b_{n+1}})\ge \min (1+\sqrt{b_0},1+\sqrt{b_{1}})$. Thus, (2) yields \[|b_{n+2}-1| \le \frac{\max (|b_n-1|, |b_{n+1}-1| )}{\min (1+\sqrt{b_0},1+\sqrt{b_{1}})} \] It means $b_n\to 1$ or $a_n\to 4$.

It is problem 2.5.37, p.53 of the book "Problems in Mathematical Anal- ysis", Vol.1 W.J.Kaczor, M.T.Novak, AMS publications see also here: http://www.math.purdue.edu/pow/spring2013/pdf/solution11.pdf

It easy to show that if for an integer $n>0$ $(4-a_n)(4-a_{n+1})\geq 0$ then $(a_k)_{k>n}$ is monotonically converging to $4$. let us discuss the other case : Assume that $a_{2k}<4$ and $a_{2k+1}>4$ for all integers $k>0$. then we easily can prove that the odd terms decrease and the even ones increase. assume there is two limits $l,g$ then : $l=\sqrt{l}+\sqrt{g}= g$. it follows that the limit is $4$.