How to prove $a_{n+1}=(1+a_n+a^2_{n-1})/3$ is a non-decreasing sequence?

Question

$$a_1=a_2=0.5$$

It isn't hard to show that $0.5\le a_n\lt 1$, and that if the sequence converge, the limit is 1.

But how to prove it's monotone?

I've tried: $$a_{n+1}-a_n=\frac{1-2a_n+a^2_{n-1}}{3}\ge 0$$ $$\frac{1+a^2_{n-1}}{2}\ge a_n$$ $$1-a_n \ge a_n - a^2_{n-1}$$

But couldn't prove any of the inequalities.

Answer 1

The hint.

By induction: $$ \begin{align} a_{n+1}-a_n&=\frac{1-2a_n+a_{n-1}^2}{3} \\&=\frac{\frac{3}{3}-2\frac{1+a_{n-1}+a^2_{n-2}}{3}+\frac{3a_{n-1}^2}{3}}{3} \\&=\frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9} \\&\geq0 \end{align} $$

Answer 2

Following the implicit advice from previous answer to use the definition twice together with induction,

$$a_{n+1}=(1+a_n+a^2_{n-1})/3$$ $$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$ $$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$

If we assume $a_n \ge a_{n-1} \ge a_{n-2}$ we get $a_{n+1} \ge a_n$