<s>[INST] How to prove the sequence $x_{n+1} = \frac{x_n}2 + \frac 1{x_n}$ is a Cauchy sequence Here is a question I do not know how to prove. Thanks for your helping！
Prove that the sequence $$x_{n+1} = \dfrac{x_n}2 + \dfrac 1{x_n}, x_0 = 1$$is a Cauchy sequence.
 [/INST] Another method: 
Firstly a bit of pedantry. The sequence is decreasing only after the second element. The first three elements of the sequence are $ 1, \frac{3}{2}, \frac{17}{12} $. 
Now consider the function $f$ which takes $ x_n \to x_{n + 1} $. 
$$ f(x) = \frac x 2 + \frac 1 x \implies f'(x)  = \frac{x^2 - 2}{ 2x^2 }$$
$ f' \gt  0  $ if $ x \gt \sqrt 2 $ and 
$ f' \lt  0  $ if $ x \lt \sqrt 2 $. So for $ x \gt \sqrt 2 $ the function $f$ is increasing. 
We will prove by induction the statement:
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; P(n)$: $\;\; \sqrt 2 \lt x_{n + 1} \lt x_n \;\;$ for $ n \ge 1 $. 
It must first be verified that $ x _1 = \frac 3 2 $ and $ x_2 = \frac{17}{12}  $ are both numbers greater than $\sqrt 2$ which can be done by squaring both sides. Then the statement is true for $n = 1$. Suppose $P(n)$ is true for arbitrary $n \in \Bbb N$. Then,
$$  \sqrt 2 \lt x_{n + 1} \lt x_{n} \implies f(x_{n + 1}) \lt f(x_{n }) \implies \sqrt 2 \lt x_{n + 2} \lt x_{n + 1} $$
which requires another small observation (to claim $ f(x_{n + 1}) \gt \sqrt 2 $) $$f(x) = \dfrac{x^2 + 2}{2x} = \dfrac{(x - \sqrt 2)^2 + 2 \sqrt 2 x}{2x} = \dfrac{(x - \sqrt 2)^2}{2x} + \sqrt 2 \gt \sqrt 2 \; \text {if} \; x \gt \sqrt 2$$
Hence $P(n)$ is true for all $n \in \Bbb N$. Hence $(x_n)$ is decreasing and is bounded below by $\sqrt 2$. Hence 
$$ (x_n) \; \text{converges} \iff (x_n) \; \text{is a Cauchy sequence} $$
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