Given that $0<a<b$, $a_0=a$, and $b_0=b$, Show that $a_{n+1}=\sqrt{a_nb_n}$ is always increasing and $b_{n+1}=\frac12(a_n+b_n)$ is decreasing

Question

This problem is very complicated and I feel like the answer should not require as much work as I am putting into it. Any tips/hints towards the solution are much appreciated!!

Answer 1

Note that $0<x<y$ implies $0<(\sqrt y-\sqrt x)^2=y+x-2\sqrt{xy} $, so that $\sqrt{xy}<\frac{x+y}2$ and also $$\tag1 0<x=\sqrt{xx}<\sqrt {xy}<\frac{x+y}2<\frac{y+y}2=y$$

By induction, we show that $$\tag2 0<a_n<a_{n+1}<b_{n+1}<b_n$$ for all $n$. Indeed, for $n=0$ we are given $0<a_0=a<b_0=b$, hence $(2)$ wth $n=0$ follows from $(1)$ with $x=a_0, y=b_0$. Now assume $(2)$ holds for $n$. Then specifically $0<a_{n+1}<b_{n+1}$ so that by $(1)$ with $x=a_{n+1}$, $y=b_{n+1}$ we obtain $(2)$ with $n+1$ in place of $n$.

We conclude that $(2)$ holds for all $n\in\mathbb N_0$. Specifically, $a_{n+1}>a_n$ for all $n$, i..e, $(a_n)$ is strictly increasing, and $b_{n+1}<b_n$, i.e., $(b_n)$ is strictly decreasing.

Answer 2

We will prove by induction that $a_n<a_{n+1}<b_{n+1}<b_n$. When $n=0$, this is the statement $$ a<\sqrt{ab}<\frac{a+b}{2}<b, $$ which follows from the AM-GM inequality (and the fact that $a<b$).

To prove it for general $n$, observe that by induction we have $a_n<b_n$. Therefore by the AM-GM inequality, $$ a_n<\sqrt{a_nb_n}<\frac{a_n+b_n}{2}<b_n. $$ Therefore $a_n<a_{n+1}<b_{n+1}<b_n$, completing the induction.