Finding the limit of a sequence If $x_1$ and $y_1$ be positive numbers, $x_{n+1}$ $=$ $1\over 2$$(x_n+y_n)$ and $2\over y_{n+1}$ $=$ $1\over x_n$ $+$ $1\over y_n$ $\forall$ $n$ $\ge$ $1$. Show that the sequences $x_n$ and $y_n$ are monotonic and converge to a common limit $l$ where $l^2$ $=$ $x_1y_1$.
I proved that these two sequences are monotone, bounded and hence convergent. But I can't find that unique limit. I've been trying it with telescopic sum, but it doesn't help at all. 
Also, I failed to establish a good relation between $x_{n+1}$ and $x_n$ or $y_{n+1}$ and $y_n$.
Please help me to find that limit.

As shown in dxiv's answer
$$
\begin{align}
x_{n+1}y_{n+1}
&=\frac{x_n+y_n}2\frac{2x_ny_n}{x_n+y_n}\\
&=x_ny_n\tag{1}
\end{align}
$$
Therefore, $x_ny_n=x_1y_1$. Additionally,
$$
\begin{align}
x_{n+1}-y_{n+1}
&=\frac{x_n+y_n}2-\frac{2x_ny_n}{x_n+y_n}\\
&=\frac{(x_n-y_n)^2}{2(x_n+y_n)}\\[3pt]
&\in\left[0,\frac{|x_n-y_n|}2\right]\tag{2}
\end{align}
$$
Thus, for $n\gt1$,
$$
0\le x_n-y_n\le\frac{|x_1-y_1|}{2^{n-1}}\tag{3}
$$
Furthermore, it follows from $(2)$, that for $n\gt1$,
$$
\begin{align}
x_{n+1}-x_n
&=\frac{x_n+y_n}2-x_n\\
&=\frac{y_n-x_n}2\\
&\le0\tag{4}
\end{align}
$$
Thus, $x_n$ is decreasing, and by $(1)$, $y_n$ is increasing.
