Using induction to show that if $x_n^2 \geq 2$, then $x_{n+1}^2 \geq 2$

Question

Let $x_1=2$. For all $n \geq 1$, define

$$x_{n+1}=\frac{1}{2} \left( x_n + \frac{2}{x_n} \right) $$

Prove by induction that $x_n^2 \geq 2$ for any $n \geq 1$.

I have a way to solve the question, which is to use AM-GM inequaity to conclude in the inductive step.

But I don't want to apply the inequality to show it. Indeed, I want to show it without using any known inequality.

I try the following in inductive step, but couldn't show what I want:

$$x_{n+1}^2 = \frac{1}{4} \left( x_n^2 + 4 + \frac{4}{x_n^2} \right) \geq \frac{1}{4} \left( 2 + 4 + \frac{4}{x_n^2} \right) = \frac{3}{2} + \frac{1}{x_n^2}$$

Answer 1

Hint: Use induction to show $x_n > 0$ always and then $$x_{n+1}=\frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \geq \sqrt{2} \Leftrightarrow \\x_n + \frac{2}{x_n} \geq 2\sqrt{2} \Leftrightarrow x_n^2 + 2 \geq 2x_n\sqrt{2} \Leftrightarrow \\ x_n^2 - 2x_n\sqrt{2} + 2 \geq 0 \Leftrightarrow \left(x_n - \sqrt{2} \right)^2 \geq 0 $$

Answer 2

The map $]0,+\infty[\to\mathbb{R},t\mapsto\frac{1}{2}\left(t+\frac{2}{t}\right)$ has an absolute minimum at $t=\sqrt 2$. The value of this minimum is $f(\sqrt 2)=\sqrt 2$. Since induction proves easily that $x_n>0$ for all $n$ (whatever $x_0>0$ is choosen), it follows that $x_n>\sqrt 2$ for all $n\ge1$.