Given $S_{1}=2$ and $S_{n+1}= \frac{S_{n}}{2} + \frac{1}{S_{n}}$. Assume $ S_{n} > 1$, show that $S_{n+1} > 1$

Question

I need help using induction on a recursive sequence.

Given $S_{1}=2$ and $S_{n+1}= \frac{S_{n}}{2} + \frac{1}{S_{n}}$

I am working on the recursive convergence to $\sqrt{2}$, therefore I want to show that it is bounded below by an arbitrary lower bound, in which I chose 1. thus by induction I want to show that $S_{n+1} > 1$, $$ S_{n} > 1$$ $$\frac{1}{S_{n}} < 1$$ $$S_n +\frac{1}{S_{n}} > ?+1$$

I get stuck here. Im not to sure how to get to my end point of $S_{n+1}$.

Answer 1

$S_n>1$

$S_n-1>0$

$(S_n-1)^2>0$

$S_n^2-2S_n+1>0$

$S_n^2+1>2S_n$

$\frac{S_n}{2}+\frac{1}{2S_n}>1$

$S_n<2S_n$

$\frac{1}{S_n}>\frac{1}{2S_n}$

$\frac{S_n}{2}+\frac{1}{S_n}>\frac{S_n}{2}+\frac{1}{2S_n}>1$

Answer 2

As for all $n$, $S_n>0$,

By AM GM inequality,

$S_{n+1}=\frac{S_n}{2}+\frac{1}{S_n}\ge2\sqrt{\frac{S_n}{2}\bullet\frac{1}{S_n}}=\sqrt2$

So $S_n\ge\sqrt2>1.$