How do you prove that a recursive sequence is increasing/decreasing?

Question

I have the two recursively-defined following sequences:

$u_0 = a$, $v_0 = b > a$, $u_{n+1} = \sqrt{u_nv_n}$, $v_{n+1} = \frac{u_n+v_n}2$.

I want to show that $u_n$ is increasing and $v_n$ is decreasing.

I was thinking that I should show it recursively. First starting off with $u_n$, we can say that $u_{1} = \sqrt{u_0v_0} = \sqrt{ab} \ge \sqrt{a^2}$, since $b>a$.

Now, I have to show that it's true for any $n$. I couldn't get anywhere from here. Am I doing this right, or is there a better way? Thanks.

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The possible duplicate is indeed the almost same question, but OP states that he/she has already solved it, and only asks about the another part of the question.

Answer 1

Hint:

First observe that for any $n$, one has $u_n<v_n$ (use the AM-GM inequality).

Supposing $b>a>0$, you easily prove $0<u_n<v_n$. So all you have to prove is

1. $\dfrac{u_{n+1}}{u_n}>1$,
2. $\dfrac{v_{n+1}}{v_n}<1$.

Answer 2

Your inductive hypothesis is $\color{red}{v_n > u_n}$ then \begin{eqnarray*} u_{n+1}=\sqrt{u_n \color{red}{v_n}} > \sqrt{u_n \color{red}{u_n}} =u_n \end{eqnarray*} and \begin{eqnarray*} v_{n+1}=\frac{\color{red}{u_n} +v_n}{2} < \frac{\color{red}{v_n}+ v_n}{2} =v_n. \end{eqnarray*} Finally by AM-GM $v_{n+1} > u_{n+1}$.