$$ a_{n+1} = \frac {2a_{n}b_{n}}{a_{n}+b_{n}}, b_{n+1}= \frac{a_{n}+b_{n}}{2}$$
where $0<a_{1}<b_{1}$ and $n \in \mathbb{N}$ and proven is that the sequence $[a_{1} b_{1}], [a_{2}, b_{2}]... $makes a nested interval and giving a number $c \in \mathbb{R}$ which is located in every interval of $[a_{n} b_{n}]$
I tried to use $b_{n+1}= \frac{a_{n}+b_{n}}{2}$ which is equal to $\frac{2b_{n+1}}{b_{n}}=a_{n} $, does it help to check the monotone of the sequences?
Note that $a_{n+1},b_{n+1}$ are the Harmonic and Arithmetic Mean of $a_n, b_n$ respectively.
Now we can use the fact here that $AM\ge GM \ge HM$ . This gives $b_n\gt a_n,\forall n\in \mathbb{N}$ (since $ b_1\gt a_1$ given)
$a_{n+1}-a_n=\frac{2a_nb_n}{a_n+b_n}-a_n$
$=\frac{a_n(b_n-a_n)}{a_n+b_n}\gt 0$ gives $a_{n+1}\gt a_n$
Again $ b_n-b_{n+1}=\frac{b_n-a_n}2\gt 0$ gives $b_{n+1}\lt b_n$
Thus clearly $[a_{n+1},b_{n+1}]\subset [a_n,b_n] \forall n\in \mathbb{N}$ and hence the intervals are nested
Oh , I forgot about the point $c$.Thanks to the comment by @Andrei
Still, I will prove why that is true.
Firstly $\lim (b_n-a_n)=0$ thus $\lim b_n=\lim a_n$
The common point is given by the above common limit.Let it be $c$
Now $a_{n+1}b_{n+1}=a_nb_n=...=a_1b_1$ can be easily checked and so
$c^2=\lim (a_nb_n)=a_1b_1$ and thus $c=\sqrt(a_1b_1)=$GM of $a_1$ and $b_1$