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:

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

  • $\begingroup$ $x_n$ or $x_n^2$ ? The title and the question mismatch ! $\endgroup$ – Nizar Jan 27 '17 at 13:31
  • 4
    $\begingroup$ $$x_{n+1}^2 - 2 = \frac{1}{4}\biggl(x_n^2 - 4 + \frac{4}{x_n^2}\biggr),$$ can you see a square on the right? $\endgroup$ – Daniel Fischer Jan 27 '17 at 13:40
  • $\begingroup$ @DanielFischer: Nice! Your method doesn't need $x_n^2 \geq 2$. $\endgroup$ – Idonknow Jan 27 '17 at 14:02

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


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 $n\ge1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.