Consider the recursively defined sequence
$a_{n+1} = \frac{1}{2}(a_{n}+\frac{A}{a_n})$ where $n \geq 0$ and $A,a_{0} > 0.$
Show that $\forall n \geq 0, a_{n+1} \geq \sqrt{A}.$
Which of the two proofs is correct?
Proof 1: Proceed by induction on $n$. Then for $n = 0$, $a_1 = \frac{1}{2}(a_{0} + \frac{A}{a_{0}}) \geq \sqrt{a_{0}(\frac{A}{a_{0}})} = \sqrt{A}$ (By AM-GM inequality). Now suppose $a_{n} \geq \sqrt{A}$. Then $$a_{n+1} = \frac{1}{2}(a_{n} + \frac{A}{a_{n}}) \geq \sqrt{a_{n}(\frac{A}{a_{n}})} = \sqrt{A}.$$
Proof 2: Let $n \geq 0$. Then $a_{n+1} = \frac{1}{2}(a_{n} + \frac{A}{a_{n}}) \geq \sqrt{a_{n}(\frac{A}{a_{n}})} = \sqrt{A}$ (By AM-GM inequality). Since $n$ is arbitrary, $a_{n+1} \geq \sqrt{A}$ $\forall n\geq 0$.
For the former, I did not use the induction hypothesis so I'm a bit doubtful.