How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$ 
Given a sequence 
  $$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
  Prove it has lower bound at $1\over 5$ and upper bound at $2$

I've tried to find a closed form for the recurrence relation, but couldn't arrive at anything. Also:
$$
x_{n+1} = \frac{x_n^4+1}{5x_n}=\frac{2(x_n^4 +1)}{2\cdot5x_n}=\frac{2}{5x_n}\cdot\frac{x_n^4+1}{2} \ge\frac{2}{5x_n}\sqrt{x_n^4\cdot1} =\frac{2x_n}{5}
$$
So I got:
$$
x_{n+1} \ge \frac{2x_n}{5} \tag1
$$
I have no ideas how to proceed. I'm not even sure it's valid to use AM-GM here. 
So my main questions are:


*

*Does this recurrence have a closed form?

*What else should I try to solve the problem?


Please note this is precalculus. I'm not allowed to use calculus.
Update
Using $(x_n^2 - 1)^2 > 0$ i get the same result as in $(1)$. Expanding the terms only shows that the sequence is greater than $0$:
$$
x_{n+1} \ge 2\cdot \left(2\over 5\right)^n
$$
which is tending to $0$ with growing $n$.
Update 2
Consider the following expressions:
$$
x_1 = 2 \\
x_2 = \frac{x_1^3}{5} + \frac{1}{5x_1} \\
\dots \\
x_{n+1} = \frac{x_n^3}{5} + \frac{1}{5x_n} \\
$$
Multiply both sides of each expression by some $z$ in the power of $n$:
$$
z\cdot x_1 = 2\cdot z \\
z^2\cdot x_2 = \left(\frac{x_1^3}{5} + \frac{1}{5x_1}\right)z^2 \\
\dots \\
z^{n+1}\cdot x_{n+1} = \left(\frac{x_n^3}{5} + \frac{1}{5x_n}\right) \cdot z^{n+1} \\
$$
Now sum them up:
$$
\sum_{k=1}^{n+1}x_k\cdot z^k = 2z + {1 \over 5}\left( \sum_{k=2}^{n+1}x_{k-1}^3z^k + \sum_{k=2}^{n+1}{z^k\over 5x_{k-1}} \right) = \\
= 2z + {1\over 5z} \left(\sum_{k=1}^{n}x_k^3z^k + \sum_{k=1}^{n}{z^k\over x_k}\right)
$$
Now define:
$$
G(z) = \sum_{k=1}^{n+1}x_k\cdot z^k
$$
From this point there may be a way to express RHS in terms of $G(z)$ but i couldn't handle that.
Update 3
This goes beyond precalculus level but anyway here is another observation inspired by @amam_Abdallah.
Define $x_{n+1} = f(x_n)$ if this function have fixed points then:
$$
\overline{x} = \frac{\overline{x}^3}{5} + \frac{1}{\overline{x}} \iff \\
\iff \overline{x} = \sqrt[^3]{5\overline{x} - {1\over \overline{x}}}
$$
This equation has two solutions:
$$
\overline{x} = \sqrt{{5\over 2} \pm {\sqrt{21} \over 2}}
$$
Perhaps this will lead someone to ideas on how to use that fact.
Update 4
Some more thoughts on the sequence:
$$
x_{n+1} = \frac{1}{5}x_n^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}x_{n-1}^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}\left(\frac{1}{5}x_{n-2}^3 + \frac{1}{5x_{n-2}} \right)^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \dots
$$
Can this somehow be wrapped into something in the form of $\prod \dots$ or $\sum \dots$?

hint
write $x_{n+1}$ as
$$x_{n+1}=\frac{x_n^3}{5}+\frac{1}{5x_n}$$
and prove it by induction.
there are two  fixed points satisfying
$$L^4-5L^2+1=0$$
$$L_1^2=\frac{5-\sqrt{21}}{2}\approx 0.7$$
$$L_2^2=\frac{5+\sqrt{21}}{2}\approx 4.7$$
