How can I prove that this sequence is monotonic? I have a sequence $(u_n)$ that is defined as:
$u_0 = 2$,
$u_{n+1} =\frac{u_n}{2} + \frac{1}{u_n}$
I have tried to prove that it is monotonic using induction but I wasn't able to succeed.  
How can I prove it easily ?  
Thank you
 A: $$
u_{n+1}-u_n=\frac{u_n}{2}+\frac{1}{u_n}-u_n=\frac{1}{u_n}-\frac{u_n}{2}
=\frac{2-u_n^2}{2u_n}
$$
So the sign is determined once we know whether $u_n^2>2$ or $u_n^2<2$.
For $n=0$, we have $u_0>\sqrt{2}$, so we can try proving $u_n>\sqrt{2}$, for every $n$.
Suppose it does for $n$. Then
$$
u_{n+1}^2-2=\left(\frac{u_n}{2}+\frac{1}{u_n}\right)^2-2=
\frac{u_n^2}{4}+1+\frac{1}{u_n^2}-2=
$$
Can you finish?
A: Let us show first that
$$u_n \ge \sqrt{2}$$
This can be done by induction: $u_1 \ge 2$, and if $u_m \ge \sqrt{2}$ then
$$u_{m+1} = \dfrac{u_m}{2}+\dfrac{1}{u_m} \geq  \sqrt{2}$$
The last inequality holds because of the following reason. Consider the function
$$f(x) = \dfrac{x}{2} + \dfrac{1}{x}$$
Its derivative:
$$f'(x) = \dfrac{1}{2} - \dfrac{1}{x^2}$$
Here we can note that if $x \ge \sqrt{2}$, then $f'(x) > 0$, which means that $f(x)$ is increasing in $[\sqrt{2}; +\infty)$. So, $f(x) \ge f(\sqrt{2}) = \dfrac{\sqrt{2}}{2} + \dfrac{1}{\sqrt{2}} = \sqrt{2}$ for $x \ge \sqrt{2}$.
So, we have just proven that $u_{m+1} = f(u_m)\ge \sqrt{2}$ (inductive step: if we assume that $u_m \ge \sqrt{2}$ then $u_{m+1} \ge \sqrt{2}$ as well). 
Now note that if $u_m \ge \sqrt{2}$ then
$$\dfrac{u_m}{2} \ge \dfrac{1}{u_m} \Rightarrow u_m \ge \dfrac{u_m}{2} +\dfrac{1}{u_m} = u_{m+1}$$
This implies that $u_{m}$ is decreasing monotonically.
