prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit

The easy way to proceed, is to show that $ b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2$ so $ b_{n} \geq a_{n} \forall n \geq 2$.
Then, $a_{n+1} = \sqrt{a_n b_n} \geq a_n$ is an monotonically increasing sequence (after $n=2$).
$b_{n+1} = \frac {1}{2} (a_n + b_n) \leq b_n$ is a monotomically decreasing sequnece (after $n=2$).
Finally, $$ b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n) \leq \frac {1}{2^n} (b_1-a_1),$$ so the difference between the sequences go to 0. Hence, these sequences converge to the same limit.

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Note: Of course we could show that since $a_i \leq b_2$, the limit of $a_i$ exists (since monotonic+bounded). But I think it's more fun to jump directly to the conclusion with the final step.

This is not an answer. Just a long comment which gives a classical formula of Gauss for this common limit.

Let $a$, $b$ be real numbers with $a<b$. We define their arithmetic-geometric mean $M(a,b)$ using the sequences $a_n$, $b_n$ defined as follows: $a_0=a$, $b_0=b$ and inductively

$$ a_{n+1}={ a_n +b_n \over 2}, \qquad b_{n+1}=\sqrt {a_n b_n }\ . $$

The two sequences converge to a unigue common limit $M(a,b)$. Gauss discovered a beautiful formula for $M(a,b)$ which can be expressed as an elliptic integral

$$ { 1 \over M(a,b)} = { 2 \over \pi} \int_0^{ { \pi \over 2}} { d \theta \over \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta}} \ . $$

If $a_n\to l$ and $b_n \to s$ then from $b_{n+1}=\frac{1}{2}(a_n+b_n)$ we get $s=\frac{1}{2}(l+s)\cdots$
To prove that they converge use arithmetic and geometric mean inequality to show (that $a_n\leq b_n$ so ...) they are monotonic ($a_n$ is increasing and $b_n$ decreasing). From $a_n\leq b_n$ conclude they are bounded.