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Given $x_1 = 2,$ and $$x_{n+1} = frac{1}{2}bigg(x_n + frac{2}{x_n} bigg),$$ show that for all $n in mathbb{N},$ $x_n geq sqrt{2}.$

I tried the following. Suppose, for contradiction, that $x_{n+1} < sqrt{2}.$ Then, $$frac{1}{2}bigg(x_n + frac{2}{x_n} bigg) < sqrt{2},$$ and $$x_n + frac{2}{x_n} < 2sqrt{2}.$$ Given that $x_n geq sqrt{2},$ it follows that $$sqrt{2} + frac{2}{x_n} leq x_n + frac{2}{x_n} < 2sqrt{2},$$ and $$sqrt{2} + frac{2}{x_n} < 2sqrt{2}.$$ So, $$frac{2}{x_n} < sqrt{2},$$ and $$sqrt{2} < x_n.$$ No contradiction. I've tried proving it directly, by squaring both sides and by leaving the expression as is, and I consistently find that the $x_n$ and the $x_n^{-1}$ are together problematic.

I also tried something that allowed me to arrive at the following: $$x_{n+1}x_n geq 2.$$ But, I don't think that's helpful.

Help?:)

$x_1geqsqrt{2}$, $x_2geqsqrt{2}$. Now suppose $x_{n-1}geqsqrt{2}$, then by AM-GM we have $x_n=frac{1}{2}(x_{n-1}+frac{2}{x_{n-1}})geqfrac{1}{2}(2sqrt{x_{n-1}frac{2}{x_{n-1}}})=sqrt{2}$

This is immediate from AM-GM: $frac{1}{2}bigg(x + frac{2}{x} bigg)$ is the arithmetic mean of $x$ and $2/x$, which is always greater than or equal to the geometric mean $sqrt{xcdotfrac{2}{x}}=sqrt{2}$ (for $x>0$).

Alternatively, if you don't know AM-GM, you can reach the same conclusion easily with a bit of calculus. Letting $f(x)=x + frac{2}{x}$, we have $f'(x)=1-frac{2}{x^2}$ which is negative for $0<x<sqrt{2}$ and positive for $x>sqrt{2}$. It follows that $f(x)$ is minimized (on $(0,infty)$) at $x=sqrt{2}$, so $frac{1}{2}f(x)geqfrac{1}{2}f(sqrt{2})=sqrt{2}$ for all $x>0$.