How to prove convergence of the Arithmetic-Geometric Mean sequence?

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Let $$a_0>b_0>0 $$and consider the infinite sequences $$\{a_n\}, \{b_n\}$$ where

$$a_{n+1}=\frac{a_n+b_n}{2}$$ and $${b_{n+1}}={(a_nb_n)}^{1/2} $$ for $n\geq0$.

Prove that the infinite sequences $\{a_n\}$, $\{b_n\}$ are convergent.

Can anyone guide me for this question? Thanks!

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First observe the followings:

$1)$: $\forall n \geq 0, a_n > 0, b_n > 0$

$2)$: $\forall n \geq 1, a_n \geq b_n$. This will lead to:

$3)$: $\forall n \geq 1, a_{n+1} < a_n$.

$4)$: $\forall n \geq 1, b_n < b_{n+1}$.

The sequence $\{a_n\}$ is decreasing, and bounded below by $0$, and the sequence $\{b_n\}$ is increasing, and bounded above by $a_0 > 0$.

Thus both sequences are convergent by Bolzano's criteria, and infact their limits are equal.

2
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BY AM-GM inequality, we can obtain that,$${b_{n+1}}=(a_nb_n)^{1/2}\le \dfrac{({a_n}+{b_n})}{2}=a_{n+1}.$$ Therefore $$b_{n+1}=(a_nb_n)^{1/2}\ge(b_nb_n)^{1/2}=b_n.$$ Hence $(b_n)$ is a increasing sequence and similarly $$a_{n+1}=\dfrac{({a_n}+{b_n})}{2}\le \dfrac{({a_n}+{a_n})}{2}=a_n$$ the sequence $(a_n)$ is a decreasing sequence. Try to show that both sequence are bdd. Then use the Monotone Convergent Theorem.