# Prove by induction that $0 < a_n < a_{n + 1} < b_{n + 1} < b_n$ for $a_{n + 1} = \sqrt{a_nb_n}$ and $b_{n + 1} = \frac 12(a_n + b_n)$.

If $a_1, b_1$ are positive numbers such that $a_1 < b_1$ and two sequences of positive numbers are defined by $a_{n + 1} = \sqrt{a_nb_n}$ and $b_{n + 1} = \frac 12(a_n + b_n)$, prove that $0 < a_n < a_{n + 1} < b_{n + 1} < b_n$.

Let $n = 1.$ Then $a_1 < b_1$ (given) and $a_2 < b_2$ by $GM \le AM.$ But how do we show $a_1 < a_2$ and $b_2 < b_1$? I tried $(b_1 - a_1) \circ (b_2 - b_1)$ where $\circ$ stands for regular addition/subtraction, but I wasn't able to deduce what I want. Any hints? Thanks.

• Hint: Both means respect order: if $y\lt z$ then $AM(x,y)\lt AM(x,z)$, etc. Feb 20, 2016 at 17:27
• $a_2=\sqrt{a_1b1} > \sqrt{a_1^2}= a_1$ since $a_1 < b_1$, $b_2 = \frac{a_1b_1}{2} < \frac{b_1, b_1}{2} = b_1$ since $a_1 < b_1$ Feb 20, 2016 at 17:32
• Very helpful info! Thanks, people. Feb 20, 2016 at 17:49

$$a_1<a_2\iff a_1<\sqrt{a_1b_1}\iff a^2_1<a_1b_1\iff a_1(b_1-a_1)>0\;\color{red}\checkmark$$
Do something similar for $\;b_n\;$