Show that the even and odd terms of the sequence described by $x_1=\frac{1}{3}, x_{n+1}=\frac{{x_n}^2+1}{5x_n}$ are monotonic and bounded. Let $(x_n)_n$ be a sequence of real numbers such that $$x_1=\dfrac{1}{3}, x_{n+1}=\dfrac{{x_n}^2+1}{5x_n}.$$ Show that the even terms are monotone decreasing and bounded and the odd terms are monotone increasing and bounded.

$\textbf{My Attempt:}$
  I first showed that $x_1<x_3$, then assumed for some positive integer $k$ that $x_{2k-1}<x_{2k+1}$. From there I attempted show that $0<x_{2k+1}-x_{2k-1}$ implies $0<x_{2k+3}-x_{2k+1}$, by substituting in $$x_{2k+3}=\dfrac{\left(\dfrac{(x_{2k+1})^2+1}{5x_{2k+1}}\right)^2+1}{5\left(\dfrac{(x_{2k+1})^2+1}{5x_{2k+1}}\right)},$$ and $$x_{2k+1}=\dfrac{\left(\dfrac{(x_{2k-1})^2+1}{5x_{2k-1}}\right)^2+1}{5\left(\dfrac{(x_{2k-1})^2+1}{5x_{2k-1}}\right)}.$$
  However, the algebra became very messy, so I think there may be a better approach that I've missed. I know the sequences of even and odd terms should both be bounded by $\frac 1 2$. I again tried to show this using induction, but ran into trouble when trying to show that $x_{2k+1}\leq \frac 1 2 \implies x_{2k+3}\leq \frac 1 2$.

Any suggestions/hints would be greatly appreciated. Thanks!
 A: As an approach, first look for a fixed point, and see how the iterates behave with respect to that point. Then show that the iterates remain within some ball around the fixed points, and then show that within this ball (or a smaller one) that the iterates converge.
Thanks to @user81146 who caught a major error in a previous proof.
A little computation shows $x_2 = \frac{2}{3}$ and $x_3 = \frac{13}{30}$.
Note that $f(x)-\frac{1}{2} = \frac{x-2}{5x} ( x -\frac{1}{2})$, and so $|f(x)-\frac{1}{2}| \le |\frac{x-2}{5x} | | x -\frac{1}{2}|$, that is,
$|x_{n+1}-\frac{1}{2} | \le |\frac{x_n-2}{5x_n} | | x_n -\frac{1}{2}|$.
Now note that $|\frac{x-2}{5x} | \le $ iff $x \ge \frac{1}{3}$ (we are only concerned with $x \ge 0$ here).
Hence if $x_n \ge \frac{1}{3}$, then $|x_{n+1}-\frac{1}{2} | \le  | x_n -\frac{1}{2}|$. To ensure that $x_n \ge \frac{1}{3}$ for all $n$, we need to have $|x_1-\frac{1}{2}| \le \frac{1}{6}$. Since $x_1 = \frac{1}{3}$, we see that this is satisfied, hence the sequence is bounded, and in particular, the distance from $\frac{1}{2}$ is non-increasing.
If we examine $\phi(x) = |\frac{x-2}{5x} |$ on $[\frac{1}{3},2]$, we see that $\phi(\frac{1}{3}) = 1$ and it is strictly decreasing. Since $x_3 > \frac{1}{3}$, we have $x_n \ge x_3 > \frac{1}{3}$ for all $n \ge 3$, and hence $\phi(x_n) \le \phi(x_3) <1$, and the above shows $|x_{n+1}-\frac{1}{2} | \le  \phi(x_3)| x_n -\frac{1}{2}|$ for all $n \ge 3$. In particular, we see that $x_n \to \frac{1}{2}$.
Note that $f$ is strictly decreasing on $(0,1)$ and $f(\frac{1}{2}) = \frac{1}{2}$. Hence if $x_n >\frac{1}{2}$ then $x_{n+1}=f(x_n) < \frac{1}{2}$ and similarly,
if $x_n <\frac{1}{2}$, then $x_{n+1}=f(x_n) > \frac{1}{2}$.
Hence the odd terms satisfy $x_n < \frac{1}{2}$ and the even terms satisfy $x_n >\frac{1}{2}$. (This also shows that $x_n \neq \frac{1}{2}$ for all $n$.)
Since we have $|x_{n+2}-\frac{1}{2}| \le \phi(x_3) |x_{n}-\frac{1}{2}|$ for $n \ge 3$, we see that
the odd terms are increasing and the even terms are decreasing.
A: $$x_{n+1} = \frac{1}{5} ( \frac{x_n}{1} + \frac{1}{x_n})$$
Let $C = 2/5,\ D=\frac{1}{5}(C+\frac{1}{C})$
So $x_1=1/3 < D$, $C<x_2 =(3+1/3)/5=2/3<1$, $x_1 < x_3 =13/30<D$ so that $$
C<x_4 <x_2,
$$
which also implies $$ D>x_5> x_3. $$ 
(1) With this observation, we use induction on odd term $$D>x_{2n+3} > x_{2n+1}$$ 
$k=3$-case is proved. If $k=2n-1$ is satisfied, that is, $x_{2n-1}<x_{2n+1}<D$, so $$(A)\ x_{2n}=\frac{1}{5} (\frac{1}{x_{2n-1}} + x_{2n-1}) > x_{2n+2} = \frac{1}{5} (\frac{1}{x_{2n+1}} + x_{2n+1}) >C
$$ So continually we have $$x_{2n+1}=\frac{1}{5} (\frac{1}{x_{2n}} + x_{2n}) < x_{2n+3} = \frac{1}{5} (\frac{1}{x_{2n+2}} + x_{2n+2}) < D.$$ 
(2) So we can conclude from odd term's increasing and bound 
that even term is decreasing and bounded (See (A)).
