If $2a_{n+2} \le a_{n+1}+a_n$, then $\lim \sup a_n \le \frac23 a_2 + \frac13 a_1$
This seems entirely straightforward.
If $b_1=a_1$, $b_2=a_2$ and $b_{n+2}=(b_n+b_{n+1})/2$ then $a_n\le b_n$. Now $b_n=\alpha+\beta(-1/2)^n$, so $b_n\to\alpha=\frac23b_2+\frac13b_1$.