How to show that this recursive sequence is monotonic?

Question

$a_1 = \frac{1}{2}, a_2 = 1, a_n = \frac{1}{2}a_{n-1} + \sqrt{a_{n-2}}$ for $n \geq 3$

I assume that $a_{n} \le a_{n+1} $ , so it means that $ \frac{1}{2}a_{n-1} + \sqrt{a_{n-2}} \le \frac{1}{2}a_{n} + \sqrt{a_{n-1}} $

Base for $n=1$ : $\frac{1}{2} = a_{1} \le a_{2} = 1$ ok.

Thesis: $a_{n+1} \le a_{n+2} $, so it means: $\frac{1}{2}a_{n} + \sqrt{a_{n-1}} \le \frac{1}{2}a_{n+2} + \sqrt{a_{n+1}}$

What should I do next? I think I should use the assumption, but I don't know how to do it.

Maybe this: $\frac{1}{2}a_{n-1} + \sqrt{a_{n-2}} \le \frac{1}{2}a_{n+2} + \sqrt{a_{n+1}} $, but it gets me nowhere. Please give me some hints.

Answer 1

Suppose $a_{k}\leq a_{k+1}$ for all $k\leq n$. Then, $$a_{n+2}=\frac{1}{2}a_{n+1}+\sqrt{a_n}\geq \frac{1}{2}a_n+\sqrt{a_{n-1}}=a_{n+1}.$$

Therefore $(a_n)$ is increasing.

Answer 2

Show the starting case for $a_1, a_2$ and $a_3$.

Now, for the induction step just write

$$a_{n+1} - a_n = \frac 12\left(\underbrace{a_n - a_{n-1}}_{\geq 0}\right) + \left(\underbrace{\sqrt{a_{n-1}} - \sqrt{a_{n-2}}}_{\geq 0}\right)$$