Trouble showing that $a_{n+1}=\frac12\left(a_n+\frac{2}{a_n}\right)$ is monotone given $a_1=2$ I've tried induction firstly by showing that $a_1>a_2$ - then assuming the expression for some $k$ I failed to make proper use of the assumption in the inductive step and I get stuck with the expression $2 < (a_n)^2$.
Next I tried directly computing both $a_{n+1}-a_n$ and $a_{n+1}/a_n$ which yielded nothing of particular use.
The most promising but failed attempt was when I tried induction on $a_{n+1}/{a_n}$ starting from the base case $a_2/a_1=3/4<1$, assuming $a_{k+1}/a_k<1$ and then making some use of the assumption in the inductive step however not being able to reach $a_{k+2}/a_{k+1} < 1$.
All of this is an attempt to show convergence of the given sequence by showing monotonicity and boundedness.
 A: Well, let's just look at your recursive formula: Let $f(x)=\frac{1}{2}(x+\frac{2}{x})$ and note that for $x>0$
$$
f(x)-x=\frac{1}{x}-\frac{x}{2}\geq 0\Longleftrightarrow \frac{1}{x}\geq \frac{x}{2} \Longleftrightarrow 1\geq \frac{x^2}{2}
$$
Thus, $f$ is increasing on $(0,\sqrt{2}]$ and decreasing on $[\sqrt{2},\infty)$. Hence, in order to check monotonicity, we should see that $f$ also respects these sets.
$$
f(x)^2-2=\frac{1}{4}\left(x^2+\frac{4}{x^2}+4\right)-2=\frac{x^2}{4}+\frac{1}{x^2}-1=\frac{1}{4x^2}(x^2-2)^2\geq 0
$$
But this implies that $f(x)$ maps all of $(0,\infty)$ into $[\sqrt{2},\infty)$. In conclusion,the algorithm will always yield a monotone, decreasing sequence after the first iteration. 
A: Observe that, AM-GM inequality gives us $ \frac{1}{2}{(a_n+2/a_n)} \geq \sqrt 2$, so $\forall n \geq 2, a_n \geq \sqrt 2$   ( here $a_1=2$ forces all $a_n$ to be positive).
Now $(a_{n+1}-a_{n})=\frac{1}{2}{(-a_n+2/a_n)} \leq 0 $ (since $a_n \geq \sqrt 2$). Thus the sequence is monotonically decreasing and the limit has to be $\sqrt 2$.
