Prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$ with $a_1=2$ is decreasing I need to prove that the sequence $a_{n+1}=0.5(a_n+\frac{1}{a_n})$, $a_1=2$ is decreasing.
I tried to use induction: I assumed that $a_n<a_n-1$, but couldn't prove that $a_{n+1}<a_n$.
How can I prove it?
 A: $a_{n+1} \geq 1$ by AM-GM inequality. Hence $a_n \geq 1$ for all $n$. Now $a_{n+1} \leq a_n$ reduces to $a_n+\frac  1 {a_n} \leq 2a_n$ or $a_n \geq \frac  1{a_n}$ which is true.
A: Note: This is a half tongue-in-cheek answer, in that it's probably not something we would expect students to come up with on their own; but hopefully it's still a useful viewpoint on the problem.
Suppose that $a_n = \coth\theta$.  Then
$$a_{n+1} = \frac{1}{2} \left( \frac{\cosh\theta}{\sinh\theta} + \frac{\sinh\theta}{\cosh\theta}\right) = \frac{\cosh^2\theta + \sinh^2\theta}{2\sinh\theta \cosh\theta} = \frac{\cosh(2\theta)}{\sinh(2\theta)} = \coth(2\theta).$$
Therefore, by induction it should be easy to show that $a_n = \coth(2^{n-1} \theta_1)$ where $\theta_1 := \operatorname{arccoth}(2)$.
Now, since $\coth$ is a decreasing function on $[0, \infty)$, and the sequence $2^{n-1} \theta_1$ is increasing, we see that $a_n$ is decreasing.
(With a bit of algebraic manipulation, the solution above could also be written as: $a_n = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1}$.)
A: This is Heron's method for calculating square roots applied for $S=1$. It's limit is $\sqrt S=1$, as seen:
$$\lambda=\frac12(\lambda+\frac 1\lambda)\implies 2\lambda=\lambda+\frac 1\lambda\implies \lambda=\frac 1 \lambda\implies \lambda=1$$
(I've ignored $\lambda=-1$ because the function is non-negative when $a_1>0$)
Now supposing $a_n=1+t$ for some $t>0$, we notice:
$$a_{n+1}=\frac12(1+t+\frac{1}{1+t})=1+(\frac{t-1}{2}+\frac{1}{1+t})<1+t=a_n$$
and so the sequence is decreasing.
A: Let's assume that $x \in I = [1, +\infty).$ It is easy to show that the image of the function $f(x) = 0.5\left(x + \frac{1}{x}\right)$ is $I$ too.
This means that for any $a_n \in I$, then also $a_{n+1} = f(a_{n}) \in I.$
In other words, since $a_1 = 2 \in I$, then $a_{n} \in I$ for all $n > 1.$
Notice that:
$$f'(x) = \frac{d f(x)}{dx} = \frac{x^2-1}{2x^2}.$$
Using this fact, it is also easy to show that $0 \leq f'(x) < 0.5$ for all $x \in I$. As a consequence $f(x)$ is Lipschitz continuous in $I$. This means that:
$$\frac{|f(x)-f(y)|}{|x-y|} \leq k ,$$
for all $x, y \in I$, for $k = \sup_{x\in I} |f'(x)| = 0.5$.
Since $k = 0.5 < 1$, then $a_{n+1} = f(a_{n})$ is a contraction mapping. For the Banach fixed-point theorem, we have that:
$$\lim_{n \to + \infty} a_{n} = a^* \in I,$$
provided that $a_1 \in I$. Here, $a^*$ is the unique fixed point of the map, at it can be evaluated as follows:
$$a_{n+1} = a_n = a^* \Rightarrow 0.5\left(a^* + \frac{1}{a^*}\right) = a^* \Rightarrow a^* = 1.$$

At this point, we know that for $a_1= 2 \in I$, then
$$\lim_{n \to + \infty} a_{n} = 1.$$
Now, observe that:
$$e_{n+1} = a_{n+1}-a_{n} = 0.5\left(\frac{1}{a_n} - a_n\right) = \frac{1-a_n^2}{2a_n^2} $$
is negative for $a_n \in I.$ Hence:
$$a_{n+1} = f(a_n) < a_n,$$
for all $n$. Since $a_{n+1} < a_{n}$ for $a_n \in I$, this prove that your sequence, which starts from $a_n =2 \in I$ is decreasing, and it will converge to $1$.
A: Update: Inspired by Daniel Schepler, I found another way to solve directly:
$$\frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n+\frac{1}{a_n}+2}{a_n+\frac{1}{a_n}-2}=\left(\frac{a_n+1}{a_n-1}\right)^2 \\\implies \frac{a_n+1}{a_n-1}=\left(\frac{a_1+1}{a_1-1}\right)^{2^{n-1}}=3^{2^{n-1}} \\ \implies 
a_n =\frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1} = 1+\frac{2}{3^{2^{n-1}}-1}$$
and it's straightforward to prove the sequence is decreasing.

Original answer
$$2a_{n+1} a_n = a_n^2+1, 2a_n a_{n-1} = a_{n-1}^2+1.\\
\implies a_n^2-a_{n-1}^2=2a_n(a_{n+1}-a_{n-1})=2a_n(a_{n+1}-a_n)+2a_n(a_n-a_{n-1})\\
\implies 2a_n (a_{n+1}-a_n)=(a_n+a_{n-1})(a_n-a_{n-1}) - 2a_n(a_n-a_{n-1})= -(a_n-a_{n-1})^2.
$$
Note that $a_n>0, \forall n$. Now you can complete your induction.
