Show that recursive sequence is decreasing 
I'm required to show that the above series is decreasing. However, I encounter a problem when I realize that in the inductive step, I have  a term for a(n) in both the numerator and denominator, which makes it difficult to show that a(n+1) > a(n+2). Any Help would be appreciated.
 A: You want to show that $a_{n+1}<a_n$ for all $n\geq1$. Writing this out yields
$$\frac{a_n}{2}+\frac{1}{a_n}<a_n,$$
and subtracting $\frac{a_n}{2}$ from both sides leaves us with
$$\frac{1}{a_n}<\frac{a_n}{2}.$$
This is equivalent to $a_n^2>2$. Can you prove this by induction?
A: If
$a_{n+1}
=\dfrac{a_n}{2}+\dfrac1{a_n}
$
then
$a_{n+1}-a_n
=-\dfrac{a_n}{2}+\dfrac1{a_n}
=\dfrac{-a_n^2+2}{2a_n}
$.
If
$a_n > \sqrt{2}$
then
$a_{n+1} < a_n$.
Also,
$a_{n+1}^2
=\dfrac{a_n^2}{4}+1+\dfrac1{a_n^2}
$
so
$a_{n+1}^2-2
=\dfrac{a_n^2}{4}-1+\dfrac1{a_n^2}
=(\dfrac{a_n}{2}-\dfrac1{a_n})^2
=(\dfrac{a_n^2-2}{2a_n})^2
$
so that
$a_{n+1} > \sqrt{2}$.
Therefore
if $a_n > \sqrt{2}$
then
$\sqrt{2} < a_{m+1} < a_m
$
for
$m \ge n$.
Therefore,
for any initial
$a_1 > 0$,
for all $n \ge 2$
we have
$\sqrt{2} < a_{n+1}
\lt a_n$
so
$a_n \to^+ \sqrt{2}$.
A: Here’s a hint. Consider when the sequence would not decrease, namely where it would stay constant. To do this, solve: 
$$x=\frac{x}{2}+\frac{1}{x}$$
$$x^2=2 \implies x=\{-\sqrt{2},\sqrt{2}\}$$
Since this sequence is always positive, the positive solution seems most relevant. As matter of fact, whenever $ a_n> \sqrt{2}$, $a_{n+1}$ decreases. I’ll leave it to you to rigorously prove the above claim, and show that $x>\sqrt{2} \implies \frac{x}{2}+\frac{1}{x}>\sqrt{2}$
