for two positive numbers $a_1 < b_1$ define recursively the sequence $a_{n+1} = \sqrt{a_nb_n}$ for two positive numbers $a_1 < b_1$ define recursively the sequence $a_{n+1} = \sqrt{a_nb_n}, b_{n+1} = \frac{a_n + b_n}{2}$. Show that $a_n, b_n$ converge to a common limit. Hint use inequality: $\sqrt{ab} \leq \frac{a+b}{2}$
attempt. 
Suppose $a_n\longrightarrow L_1$ and $b_n \longrightarrow L_2$
using the hint:
$$\lim_{n \to \infty} 4a_nb_n < \lim_{n \to \infty} (a_n + b_n)^2$$
by properties of limits:
$$4(\lim_{n \to \infty} a_n)(\lim_{n \to \infty} b_n) <(\lim_{n \to \infty}a_n +\lim_{n \to \infty}b_n)^2$$
$$4L_1L_2 \leq (L_1 + L_2)^2$$ 
$$0 \leq (L_1 - L_2)$$ after taking the square root from each side. 
Then this implies
$$L_2 \leq L_1$$
Since there is an equality in the less than equal to expression then the limits can be equal and as such they converge to a common limit.
I didn't have any ideas of what else to try hopefully I was not reckless with my attempt.
 A: The inequality $\sqrt{ab}\leq \frac{a+b}{2}$ implies that $a_n\leq b_n$ for all $n$. This in turn implies that
$$ b_{n+1}=\frac{a_n+b_n}{2}\leq b_n $$
Therefore $\{b_n\}$ is a decreasing sequence of positive real numbers, hence has a limit b.
Next,
$$ a_{n+1}=\sqrt{a_nb_n}\geq \sqrt{a_n^2}=a_n $$
Therefore $\{a_n\}$ is an increasing sequence. It is bounded above because $a_n\leq b_n$ for all $n$ and $\{b_n\}$ is bounded, so it has a limit $a$.
Finally, take $n\to\infty$ in both sides of $b_{n+1}=\frac{a_n+b_n}{2}$ to obtain $b=\frac{a+b}{2}$, or $a=b$.
A: By the hint we get $a_n\leq b_n$ for every $n\in\mathbb{N}$, we also see that $a_n$ is monotonous increasing and $b_n$ is decreasing, so both sequences are monotonous and bounded $a_n\leq b_1$, $b_n\geq a_1$ and so they are convergent. Now we get by properties of limit
$$\lim_{n\to\infty}b_n=\lim_{n\to\infty} b_{n+1}=\lim_{n\to\infty}\frac{a_n+b_n}{2}=\frac{1}{2}\left(\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n\right)$$ and so
$$\frac{1}{2}\lim_{n\to\infty} a_n=\frac{1}{2}\lim_{n\to\infty} b_n $$
and the limits are the same.
