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I want to show that $x_1=2,x_{n+1}=\frac{x_n^2+2}{2x_n}$ is a convergent sequence and thus want to bound it below by $\sqrt{2}$. I tried to use induction but my problem is that if $x_n>\sqrt{2}$ then the denominator is greater than 4 but the denominator is less than $\frac{1}{2\sqrt{2}}$ so they compete against each other. How can solve this issue?

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You can solve $$x_{n+1}=\frac{x_n^2+2}{2x_n}>\sqrt2\\x_n^2+2>2\sqrt2 x_n\\x_n^2-2\sqrt2 x_n+2=(x_n-\sqrt2)^2>0$$ Which is true for any $x_n\neq \sqrt2$

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  • $\begingroup$ Though of course you need to assume $x_n>0$ for all positive integer $n$. $\endgroup$ – Ahmed S. Attaalla May 17 '17 at 21:32
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We have,

$$x_n>0$$

Because positive numbers are closed under multiplication and addition. As such, we have,

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

By the Am-GM inequality.

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