Bound on subsequent terms of inductively defined sequence Let $(a_n)$ be the sequence defined by $a_0=1$ and $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$. Show that $|a_{n+1}-a_n|\leq(\frac{1}{2})^{n+1}$.
Using the induction hypothesis I can show that
$$|a_{n+1}-a_n|\leq \frac{|a_n-a_{n-1}|}{2} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|\leq (\frac{1}{2})^{n+1} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|.$$
I cannot get rid off the extra term and I have tried different approaches, but this one seems like the most promissing one because the bound appears naturally. I think I am missing something obvious.
Thank you.

Assume you want $|a_{n+1}-a_n| \le \frac{1}{2^{n+1}}$
First $a_n\ge 0$. If $n\ge 1$ then $a_n \ge 2\sqrt{\frac{a_{n-1}}{2} \cdot \frac{1}{a_{n-1}}} = 2\sqrt{\frac 12}=\sqrt 2$ via AM-GM.
$a_{n+1} - a_n = \frac{1}{a_n} - \frac{a_n}{2}=\frac{2-a_n^2}{2a_n} \le 0$
Lastly,
$$a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}, a_n=\frac{a_{n-1}}{2} + \frac{1}{a_{n-1}}\\
\implies a_{n+1} - a_n = (a_n-a_{n-1}) \left( \frac 12 - \frac{1}{a_n a_{n-1}}\right) \\
\implies |a_{n+1} - a_n | \le \frac 12|a_n-a_{n-1} | \le \frac{1}{2^n}|a_1-a_0|=\frac{1}{2^{n+1}}$$
Note: we have $0 \le \frac 12 - \frac{1}{a_n a_{n-1}} \le \frac 12$. LHS because $a_n\ge \sqrt 2 $, RHS because $a_n>0$
