For two numbers $a$ and $b$ with $0 \leq a\leq b $ we have that:
$a_{n+1} =\sqrt{a_nb_n} \quad$ , $\qquad b_{n+1} = \frac {a_n+b_n}2 \quad,$ $\qquad a_0 = a, \quad b_0=b $
And I have to prove for $n\ge1$ that $b_n-a_n < (\frac {1}2)^n(b-a) \quad $ if $a\neq b$
In previous exercises I've proved that:
(i) $\sqrt{ab} \leq \frac{a+b}2$ $\quad$ for $\; 0\leq a \leq b \;$ with equality if and only if $a=b$
(ii) $ a <\sqrt{ab} \; $ and $\frac{a+b}2 <b \; $ for $0 \leq a<b$
By using (i) and (ii) I have done this so far:
$b_n-a_n =\frac{a_{n-1}+b_{n-1}}2 - \sqrt{a_{n-1}b_{n-1}} $
$\qquad \quad < \frac{a_{n-1}+b_{n-1}}2-a_{n-1}=\frac{1}2(b_{n-1}-a_{n-1}) \quad$ for $n\ge1$
I'm stuck from here.
(Edited)