prove two dependent sequences converge

Further developing my comment, if we denote by $\mbox{AM}$ the arithmetic mean and by $\mbox{HM}$ harmonic mean we have

$$b_{n+1} = \mbox{AM}(a_n,b_n)\tag{1}\label{1}$$ and $$a_{n+1} = 2\mbox{AM}(a_n,b_n)-\mbox{HM}(a_n,b_n)\geq \mbox{AM}(a_n,b_n)=b_{n+1},$$ with equality only if $a_n=b_n$.

So, considering that $0<b_1<a_1$, for all $n\in \Bbb Z^+$ we have $$b_n<a_n\tag{2}\label{2}.$$

Condition \eqref{2} guarantees, together with \eqref{1}, that $$0<b_n<b_{n+1}<a_{n+1}.\tag{3}\label{3}$$

Furthemore we have \begin{eqnarray} a_{n+1}-a_n &=& a_n+b_n - \frac{2a_nb_n}{a_n+b_n} -a_n=\\ &=&\frac{b_n^2-a_nb_n}{a_n+b_n}=\\ &=&\frac{b_n(b_n-a_n)}{a_n+b_n}<0. \end{eqnarray} Thus $a_{n+1}<a_n$, which together with \eqref{3} leads to $$0<b_1<b_2<\cdots<b_{n}<\cdots<a_n<\cdots<a_2<a_1.$$

In conclusion

1. $(a_n)$ is monotonically decreasing and bounded from below by $b_1$, thus it converges to some $\alpha$.
2. $(b_n)$ is monotonically increasing and bounded from above by $a_1$, thus it converges to some $\beta \leq \alpha$.
3. Suppose $\beta < \alpha=\beta + \Delta$, and take $n$ large enough so that $\beta-\frac{\Delta}2<b_n$. Then $$b_{n+1} = \frac{a_n+b_n}2>\frac{\alpha +\beta-\frac{\Delta}2}2>\beta,$$ which is a contradiction. So it must be $\alpha = \beta$. $\blacksquare$

You may turn it into a fixpoint problem as follows:

• Consider $q_n = \frac{b_n}{a_n} \Rightarrow 0<q_1 <1$ and $a_1 >0$ and you have

$$a_{n+1} = a_n\frac{1+q_n^2}{1+q_n} \mbox{ and } q_{n+1} = \frac{1}{2}\frac{(1+q_n)^2}{1+q_n^2}$$

Now, it is enough to show that $\boxed{\lim_{n\to \infty}q_n = 1}$.

To do so, consider $f(x) = \frac{1}{2}\frac{(1+x)^2}{1+x^2}$ and check the following properties:

• $f:[0,1] \Rightarrow [\frac{1}{2},1]$, because for $x \in [0,1]$ you have $$f(x) = \frac{1}{2}\left(1+\frac{2x}{1+x^2} \right)\leq \frac{1}{2}\left(1+\frac{1+x^2}{1+x^2} \right)=1$$ $$f(x)=\frac{1}{2}\frac{(1+x)^2}{1+x^2}\geq \frac{1}{2}\frac{(1+x)^2}{1+x}=\frac{1}{2}(1+x)\geq \frac 12$$
• $f$ has the fixpoint $x=1$ since $f(1) = 1$.
• $f'(x) = \frac{1-x^2}{(1+x^2)^2}$ and for $x \in [\frac{1}{2},1]$ you have $$0\leq f'(x) = \frac{1-x^2}{(1+x^2)^2} \leq 1-x^2 \leq \frac 34$$

So, $f$ is a contraction on $[\frac 12, 1]$ and you can conclude that $q_n$ converges to the fixpoint $x=1$.

Since $a_{n+1} = a_n \frac{1+q_n^2}{1+q_n} \stackrel{q_n \in [0,1]}{\leq}a_n \frac{1+q_n}{1+q_n}=a_n$, you see that $a_n$ is decreasing and bounded from below (by $0$). So, $a_n$ is convergent and hence $b_n$, too and they must have the same limit.