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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!

Silent
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    AM-GM on the stuff in the bracket? – Shraddheya Shendre Jan 16 '17 at 18:11
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    To show that $$x+\frac2x\geqslant2\sqrt2$$ for every $x>0$ (which implies your result), an option is to differentiate the LHS and check that it is minimum at $x=2\sqrt2$. Another option is to note that $$x+\frac2x-2\sqrt2=\left(\sqrt{x}-\sqrt{\frac2x}\right)^2\geqslant0$$ And I am sure there are tons of other approaches... – Did Jan 16 '17 at 18:15
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    Let $,y_n = x_n / \sqrt{2},$ then the recurrence reduces to a form with the easily recognizable $a+\frac{1}{a} \ge 2,$: $$y_{n+1} = \cfrac{1}{2}\left(y_n + \cfrac{1}{y_n}\right) ;\ge; 1$$ – dxiv Jan 16 '17 at 21:18

2 Answers2

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By AM-GM

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

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

DanielWainfleet
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