# monotonicity of geometric and arithmetic mean

I have to proof the following thing and I am completely stuck.: Let $a=a_0\in\mathbb{R}$ and $b=b_0\in\mathbb{R}$ be postive and $a<b$.

Let $a_{n+1}=\sqrt{(a_n b_n)}$ and $b_{n+1}=\frac{a_n +b_n}{2}$.

Proof that: $a_{n+1}\geq a_n$ and $b_n\geq b_{n+1}$.

I know that $\sqrt{{{(a_n b_n)}}}\leq\frac{a_n +b_n}{2}$.

If I use this formula it is quite easy to show that $a_{n+1}\geq a_n$ and $b_n\geq b_{n+1}$. But is that enough? Don't I have to give a proof using induction? I would be very grateful for any help! Thank you very much in advance.

• Induction is not a must. You could use any method as long as it is acceptable mathematically. – Jlamprong May 14 '14 at 19:23
• But I still dont see how using the formula proofs it for all n. – Hans May 14 '14 at 19:30

That $a_n > 0$ and $b_n > 0$ can be assumed to be true or proven from their respective recursive formulas. $a_{n+1} = \sqrt{a_nb_n} \geq \sqrt{a_n^2} = a_n$, because $b_n \geq a_n$, and the latter inequality is true due to the AM-GM inequality. $b_n - b_{n+1} = \dfrac{b_n - a_n}{2} \geq 0$ since $b_n \geq a_n$ is true also due to AM-GM inequality.
Note: $b_n = \dfrac{a_{n-1} + b_{n-1}}{2} \geq \sqrt{a_{n-1}b_{n-1}} = a_n$