Show $a_{n+1} \geq \sqrt{B}$

Question

Suppose $a_0$ and $B$ are positive numbers , define a sequence $a_{n+1} = \frac{1}{2}(a_n +B/a_n)$. Show that $a_{n+1} \geq\sqrt B$.

Answer 1

This is just the AM-GM inequality: $$ a_{n+1}=\frac{1}{2}\Big(a_n+\frac{B}{a_n}\Big)\geq\sqrt{a_nB/a_n}=\sqrt{B}. $$ For strict rigor, you should make a quick induction argument to show that $a_n>0$ for all $n$. To see that $a_{n+1}\geq B$ is tight, see what happens when $a_0=\sqrt{B}$.