Help with induction on a recursive sequence.

Question

I'm currently working on this problem:

At first, this looked like a pretty straightforward induction problem. But, once I started working on (b), I ran into an issue.

I can show that my base case is greater than or equal to 2, and then I assume for some n in N that x sub n is greater than or equal to 2. Next, I want to show that x sub n+1 is greater than or equal to 2, and I planned to do so with a simple series of inequalities. However, the 1/(x sub n) in the recursive definition is causing difficulty for me because it breaks my inequality.

Can someone give me some pointers here? Thanks!

Answer 1

maybe you don't need induction. Maybe if you recall the AM-GM inequality

$$ \frac{a+b}{2} \geq \sqrt{ab} $$

For positive $a,b$. Now, as for your problem, notice that

$$ x_{n+1}^2 = \frac{x_n^2}{4} + 1 + \frac{1}{x_n^2} $$

Now, with $a= \frac{x_n^2}{4}$ and $b = \frac{1}{x_n^2}$, one has

$$ x_{n+1}^2 \geq 2 \sqrt{ \frac{x_n^2}{4}\frac{1}{x_n^2} } + 1 = 1+1=2$$

Answer 2

$(\frac 12 x_n + \frac 1{x_n})^2 = \frac 14 x_n^2 + \frac 1{x_n}^2 + 1$

If $x_n^2 = 2$ then $\frac 14 x_n^2 + \frac 1{x_n}^2 + 1=2$. But if $x_n^2 > 2$ then

Let $x_n^2 - 2 = \epsilon > 0$

Then $\frac 12 - \frac 1{x_n}^2 = \frac 12 - \frac 1{2+ \epsilon} = \frac {2+\epsilon - 2}{2(2+\epsilon)}=\frac {\epsilon}{4+ 2\epsilon}$

So $\frac 14 x_n^2 + \frac 1{x_n}^2 + 1 = \frac 14(2 + \epsilon) + \frac 12 - \frac {\epsilon}{4+ 2\epsilon}+1=2 + \frac \epsilon 4 - \frac \epsilon {4+ 2\epsilon} > 2 +\frac \epsilon 4 - \frac \epsilon 4=2$

....

In general if $1 \le a < b$ then $b-a > \frac 1b - \frac 1a$. [because $\frac 1b - \frac 1a = \frac {a-b}{ab}< \frac {a-b}1$]. A handy fact to know.