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!

$$x_{n+1} = \frac{1}{5} ( \frac{x_n}{1} + \frac{1}{x_n})$$
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)).
