Natural sequence 
Let $a_n$ be a natural sequence such that $a_{n+2}=\left\lfloor\frac{2a_{n+1}}{a_n}\right\rfloor +\left\lfloor\frac{2a_n}{a_{n+1}}\right\rfloor $. Show that there exists integer $k$ such that $a_k=4$ and $a_{k+1}\in\{3,4\}$. 

I'm trying to prove this using a "proof by cases" but it is very complicated. 
 A: This is an old contest problem. It was the 6th problem of the Iran 2nd round competition in 2013. You can find something more about this question here. Anyway here's my proof:
Lemma 1: For every $n\ge 3$ we have $a_n \ge 3$
Assume the opposite an let $a_n<3$ for some $n\ge 3$. If $a_{n-1} \ge a_{n-2}$, then:
$$\frac{2a_{n-1}}{a_{n-2}} \ge 2 \implies \left[\frac{2a_{n-1}}{a_{n-2}} \right] \ge 2 \implies \left[\frac{2a_{n-2}}{a_{n-1}} \right] = 0 \implies a_{n-1} > 2a_{n-2} \implies \frac{2a_{n-1}}{a_{n-2}} \ge 4$$
This contradicts that $a_n<3$. Similarly we discard the case when $a_{n-2}\ge a_{n-1}$
Lemma 2: For every $n\ge3$ we have $a_{n+1}=a_n$ or $a_{n+2} < \text{max}( a_n,a_{n+1})$
Assume that $a_{n+1} = a_n$. If $a_n = \text{max}(a_n,a_{n+1})$, then $\frac{2a_{n+1}}{a_n} < 2$. On the other side:
$$a_{n+1}a_n \ge 3 \implies \frac{2a_n}{a_{n+1}} \le \frac{2a_n}{3} \implies a_{n+2} = \left[\frac{2a_{n+1}}{a_{n}} \right] + \left[\frac{2a_{n}}{a_{n+1}} \right] \le 1 + \frac{2a_n}{3} \le \frac{3a_n}{3} = a_n$$
But now if $a_{n+2} \ge \text{max}( a_n,a_{n+1})=a_n$ using the previous inequality we have that $a_{n+1} = a_n = 3$. A contradiction. Similarly we discard the case when $a_{n+1} = \text{max}(a_n,a_{n+1})$ in order to prove the claim.
Lemma 3: There exists $k \in \mathbb{N}$ s.t. $a_k = a_{k+1}$
Assume the opposite. Then from the previous lemma we have that: $a_{n+1} < \text{max}( a_n,a_{n-1})$ and $a_{n+2} < \text{max}( a_n,a_{n+1}) \le \text{max}( a_n,a_{n-1})$
Hence $\text{max}( a_{n+1},a_{n+2}) < \text{max}( a_n,a_{n-1})$. This implis that the function $\text{max}( a_n,a_{n+1})$ has a decreasing subsequence, which is impossible in the natural numbers. Hence the lemma is proven
Proof:
Now using the Lemma 3 we have that there exists $k \in \mathbb{N}$ s.t. $a_k = a_{k+1}$. This implies that $a_{k+2} = 4$. If $a_k = a_{k+1} \in \{3,4\}$, then the claim is proven. Now assume the opposite and let $a_k = a_{k+1} > 4$. Then applying the Lemma 3 again we have that there exists $m \in \mathbb{N}$ s.t. $a_{k+m+1} = a_{k+m}$. Now using the thing we established during the proof of Lemma 3 we have:
If $m$ is even:
$$a_{k+1} = \text{max}( a_{k+1},a_{k+2}=4) > \text{max}( a_{k+3},a_{k+4}) > ... > \text{max}( a_{k+m},a_{k+m+1}) = a_{k+m}$$
If $m$ is odd:
$$a_{k+1} = \text{max}( a_{k+1},a_{k+2}=4) > \text{max}( a_{k+3},a_{k+4}) > ... > \text{max}( a_{k+m-1},a_{k+m}) \ge a_{k+m}$$
This means that if we have two consecutive equal numbers greater than $4$, eventually we will have two consecutive equal numbers, which are less than the previous ones. After finitely many steps we will find some consecutive equal numbers equal or less than 4. This is enough as the very next two elements of the sequence are $4,4$ or $4,3$
