Estimate of the difference between arithmetic and harmonic mean. Let $0<a<b$
The excercise is to Show that the nested interval defined recursively $I_1:[a,b],I_{n+1}:[a_{n+1},b_{n+1}]$ with $a_{n+1}=\frac{2a_nb_n}{a_n+b_n}$ and $b_{n+1}=\frac{a_n+b_n}{2}$ is actually a nested interval and then that the number which is always in all intervals is $\sqrt{ab}$. 
I Need some help to Show that the distance of the intervals goes to Zero.
When I look at the difference $b-a$ I get
with some manipulations the fraction 
$b-a=\frac{(a-b)^2}{2(a+b)}$
What I am Looking for is a smaller estimate that depends on $b$ and $a$. Can you help me please? 
 A: Straighforward algebra
with a little cleverness.
If $0 < a_n < b_n$,
then
$\begin{array}\\
a_{n+1}-a_n
&=\frac{2a_nb_n}{a_n+b_n}-a_n\\
&=\frac{2a_nb_n-a_n(a_n+b_n)}{a_n+b_n}\\
&=\frac{a_nb_n-a_n^2}{a_n+b_n}\\
&=\frac{a_n(b_n-a_n)}{a_n+b_n}\\
&\gt 0\\
b_{n+1}-b_n
&=\frac{a_n+b_n}{2}-a_n\\
&=\frac{a_n+b_n-2b_n}{2}\\
&=\frac{a_n-b_n}{2}\\
&\lt 0\\
b_{n+1}-a_{n+1}
&=\frac{a_n+b_n}{2}-\frac{2a_nb_n}{a_n+b_n}\\
&=\frac{(a_n+b_n)^2-4a_nb_n}{2(a_n+b_n)}\\
&=\frac{(a_n-b_n)^2}{2(a_n+b_n)}\\
&\gt 0\\
\end{array}
$
so that
$a_n < a_{n+1}
\lt b_{n+1} < b_n
$.
Also,
$\begin{array}\\
b_{n+1}-a_{n+1}
&=\frac{(a_n-b_n)^2}{2(a_n+b_n)}\\
&=\frac{(b_n-a_n)(1-a_n/b_n)}{2(a_n/b_n+1)}\\
&<\frac{b_n-a_n}{2}
\quad\text{since }1-a_n/b_n<1, a_n/b_n+1>1\\
\end{array}
$
Therefore
$0 < b_{n+k}-a_{n+k}
\lt \frac{b_n-a_n}{2^k}
\to 0
$.
A: Hints:
$$I_1:[a,b],I_{n+1}:[a_{n+1},b_{n+1}] \Rightarrow I_1:[a_1,b_1].$$
And based on HM-GM-AM:
$$a_1\le a_2=\frac{2}{\frac1{a_1}+\frac1{b_1}}\le \sqrt{a_1b_1}\le \frac{a_1+b_1}{2}=b_2 \le b_1;\\
a_2\le a_3=\frac{2}{\frac1{a_2}+\frac1{b_2}}\le\sqrt{a_2b_2}\le \frac{a_2+b_2}{2}=b_3\le b_2;\\
\vdots\\
a_n\le a_{n+1}=\frac{2}{\frac1{a_n}+\frac1{b_n}}\le \sqrt{a_nb_n}\le \frac{a_n+b_n}{2}=b_{n+1}\le b_n.$$
They are the bounded monotonic sequences. 
