\displaystyle a_{n + 3} = a_{n + 2}^2 + \frac{1}{6} a_{n + 1} + \frac{1}{9} \geq a_{n + 1}^2 + \frac{1}{6} a_{n + 1} + \frac{1}{9} \geq a_{n + 1}^2 + \frac{1}{6} a_n + \frac{1}{9} = a_{n + 2}