Proving two sequences converge to the same limit $a_{n+1}\frac{a_n+b_n}{2} \ , \ b_{n+1}=\frac {2a_nb_n}{a_n+b_n} $ 
$\text{We have two sequences}$ $(a_n), (b_n)$ where $0<b_1<a_1$ and: 
$$a_{n+1}=\frac{a_n+b_n}{2} \ , \ b_{n+1}=\frac {2a_nb_n}{a_n+b_n} $$
Prove both sequences converge to the same limit and try to find the
  limit.

What I did: Suppose $\displaystyle\lim_{n\to\infty}a_n=a, \displaystyle\lim_{n\to\infty}b_n=b$ So $\displaystyle\lim_{n\to\infty} \frac {a_n+b_n} 2= \frac{a+b} 2 =K$
Take $a_{n+2}= \frac {a_{n+1}+b_{n+1}}{2}=\frac {\frac{a_n+b_n}{2}+\frac {2a_nb_n}{a_n+b_n}}{2}=...=X$
We know that as $n$ tends to infinity $\lim x_n= \lim x_{n+1}$ so: $X=K$ and after some algebra I get $a=b$
As for the limit, it depends on only one of the sequences, since both tend to the same limit. The limit can be any constant or $\pm\infty$. 
Is this approach correct ?
I excluded the algebra because I type this manually and to make the solution easier to read. 

*

*Just show via an induction that $b_n\le b_{n+1} \le a_{n+1} \le a_n$: this proves that both sequences are convergent.

*Then take the limit in the definition and the previous inequality: you get
$$A = \frac 12 (A+B)
\\A\ge B$$so $A=B.$


details for 1.:
a) The inequality 
$$
u<v\implies \frac {u+v}2<v
$$is trivial.
b)$$
u<v\implies \frac 1u > \frac 1v
\\ \implies \frac 1u > \frac 12 \left(\frac 1u +\frac 1v\right)
=\frac{u+v}{2uv}\implies u<  \frac{2uv}{u+v} 
$$
c) As $0\le(\sqrt{u}-\sqrt{v})^2$,
$$
\sqrt{uv}\le \frac{u+v}2\\
4uv\le (u+v)^2\\ 
\frac{2uv}{u+v} \le  \frac {u+v}2
$$
