i can Just prove that there exist suchan m so that a_m=4 first it's obviouse that there are infinitly many naturals i such that a_i<a_i+1 just do soma case works like : 2(a_i) < a_i+1....then from this get a_i+2,a_i+3 < .5 (a_i+1) the case a_i <a_i+1 <2 a_i is easier the rest of cases is when a_i=1(obviouse) ,a_i=2("too) and a_i=3 the equality cases of above inequalities are easy to arrive with the last case.

Here it is a solution which is not my own . For $n\ge 3$ always $a_n\ge 2$. Let $a_n=a\ge 2$, $a_{n+1}=b\ge 2$. Then $a_{n+2}=f(c),c=\frac{a_{n+1}}{a_n}, f(c)=[\frac 2c]+[2c]$. $a_{n+2}=max(a_n,a_{n+1}$ only if one of $a_n,a_{n+1}$ equal $2$. If $a_n=2$, $a_{n+1}=a>4$, then $a_{n+2}=a_{n+1}\to a_{n+3}=4 a_{n+4}=[\frac a2]+[\frac 8a]<a$. It mean if $max(a_n,a_{n+1})=a>4$, then $max(a_{n+4},a_{n+5})<a$, therefore for $n>N$ always $max(a_n,a_{n+1})\le 4$. $a_n=2,a_{n+1}=2\to a_{n+2}=4, a_{n+3}=5,a_{n+4}=3,a_{n+5}=4,a_{n+2k}=3,a_{n+2k+1}=4$. By analogy check $a_n=2,a_{n+1}=3, a_n=3,a_{n+1}=3, a_{n}=3,a_{n+1}=4, a_n=a_{n+1}=4$.