How to show that for sequence $x_1=2$, and $x_{n+1}=\frac12\big(x_n+\frac2{x_n}\big)$; for all $n$, $(x_n)^2>2$?

Question

How to show that for sequence $x_1=2$, and $x_{n+1}=\frac12\big(x_n+\frac2{x_n}\big)$; for all $n$, $(x_n)^2>2$ ?

I tried using induction: let $(x_k)^2>2$, so $(x_{k+1})^2=\frac14\big(x_k^2+\frac4{x_k^2}+4\big)>\frac32+\frac1{x_k^2}$. I can't go further!

Answer 1

Let $x_n^2=2+d_n .$ We have $x_{n+1}=(x_n+2/x_n)/2=(x_n^2+2)/2x_n=(4+d_n)/2x_n. $ Therefore $$ d_{n+1}=x_{n+1}^2-2=\frac {(4+d_n)^2}{4x_n^2}-2=\frac {(4+d_n)^2}{4(2+d_n)}-2=$$ $$=\frac {(16+8 d_n+d_n^2)-8(2+d_n)}{4(2+d_n)}=\frac {d_n^2}{4(2+d_n)}=\frac {d_n^2}{4x_n^2}=\left(\frac {d_n}{2x_n}\right)^2.$$.

Answer 2

By AM-GM

$$x_n+\frac{2}{x_n}\ge 2\sqrt{\left(x_n\cdot \frac{2}{x_n}\right)}=2\sqrt{2} \Rightarrow x_{n+1}=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)\ge \sqrt{2} \Rightarrow (x_{n+1})^2>2$$