We'll rewrite this as \[(a_{n+1})^2 + (a_{n+2} - 1)(a_n - 1) \leq 1\] If both \(a_n\) and \(a_{n+2}\) are \(\le 1\), then \(a_{n+1} \le 1\). Similarly, if both \(a_n\) and \(a_{n+2}\) are \(\ge 1\) then \(a_{n+1} \le 1\). Oh interesting, we get \[(a_{n+2} - 1)(a_n - 1) \leq (1 - a_{n+1})(1 + a_{n+1})\] Indeed, as we could have gotten before, if \(a_{n+1} \ge 1\) then exactly one of \(a_n\) and \(a_{n+2}\) must be \(\le 1\). So suppose we have a sequence starting with: \[GGL : GGLGGLGGL\dots\] \[LGG : LGGLGGLGG\dots\] \[GLG : GLGGLGGLG\dots\] \[LLL : LLLLLLLLL\dots\] So everything possible is periodic every three turns, and we should prove that the sequences \(LGG\) and \(GLG\) are somehow not possible. However, \(LGG\) and \(GLG\) are isomorphic to \(GGL\), just delayed by \(1\) or \(2\). So what's going on? It seems like we must show that everything must be \(\le 1\).

After some thinking, it seems like looking very locally at what is wrong is not working. So I'll try to add or subtract some conditions (which actually looks nice). \[a_{n+1}(a_{n+1} + a_{n+3}) + a_{n+2}(a_{n} + a_{n+2}) \le (a_{n+1} + a_{n+3}) + (a_n + a_{n+2})\] Bruh okay so of course \(a_{n+1}\) and \(a_{n+2}\) can't be both \(G\). And since everything with a single \(G\) definitely has adjacent \(G\)s as seen above, we are done.