If $2a_{n+2} \le a_{n+1}+a_n$, then $\lim \sup a_n \le \frac23 a_2 + \frac13 a_1$

Question

This is a reformulation of a deleted question:

If $a_1 > 0$ and $a_2 > 0$ and $2a_{n+2} \le a_{n+1}+a_n$, show that $\lim \sup a_n \le \frac23 a_2 + \frac13 a_1$.

My proof involves showing that $a_{n+2} \le u_{n} a_{n+1} + (1-u_n)a_n $ where $u_n \to \frac23$, and I wondered if there is a simpler proof.

I will post my proof in a couple of days if no one does.

Answer 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 $b_n\to\alpha=\frac23b_2+\frac13b_1$.