# Prove all elements $x_n \geq \sqrt{2}.$ [duplicate]

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 $$x_n + \frac{2}{x_n} < 2\sqrt{2}.$$ Given that $$x_n \geq \sqrt{2},$$ it follows that $$\sqrt{2} + \frac{2}{x_n} \leq x_n + \frac{2}{x_n} < 2\sqrt{2},$$ and $$\sqrt{2} + \frac{2}{x_n} < 2\sqrt{2}.$$ 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?:)

## marked as duplicate by Martin R, Community♦Jan 10 at 6:06

• Just apply "AM >= GM" to $\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg)$ ... – Martin R Jan 10 at 6:02
• AM-GM inequality. – xbh Jan 10 at 6:02
• What is the AM-GM inequality? – Rafael Vergnaud Jan 10 at 6:02
• – Martin R Jan 10 at 6:02
• Got it! Thanks :) – Rafael Vergnaud Jan 10 at 6:04

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 $$\sqrt{x\cdot\frac{2}{x}}=\sqrt{2}$$ (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 $$0 and positive for $$x>\sqrt{2}$$. It follows that $$f(x)$$ is minimized (on $$(0,\infty)$$) at $$x=\sqrt{2}$$, so $$\frac{1}{2}f(x)\geq\frac{1}{2}f(\sqrt{2})=\sqrt{2}$$ for all $$x>0$$.
$$x_1\geq\sqrt{2}$$, $$x_2\geq\sqrt{2}$$. Now suppose $$x_{n-1}\geq\sqrt{2}$$, then by AM-GM we have $$x_n=\frac{1}{2}(x_{n-1}+\frac{2}{x_{n-1}})\geq\frac{1}{2}(2\sqrt{x_{n-1}\frac{2}{x_{n-1}}})=\sqrt{2}$$