How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$

Question

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:

1. Does this recurrence have a closed form?
2. 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 $\sum \dots$?

Answer 1

Over the interval $\left[\frac{1}{5},2\right]$ the function $f(x)=\frac{x^4+1}{5x}$ has an absolute minimum occurring at $x=3^{-1/4}$; if we take $I=\left[\frac{4}{5\cdot 3^{3/4}},2\right]$ we have $f(I)\subset I$. $f(2)=\frac{17}{10}$ and $f\circ f$ (but not $f$!) turns out to be a contraction of the metric space $J=\left[\frac{4}{5\cdot 3^{3/4}},\frac{17}{10}\right]$ since $f(f(J))\subset J$ and $\left|\frac{d}{dx}f(f(x))\right|\leq 0.94$ over $J$. By the Banach fixed point theorem, $x_n$ converges to the only fixed point of $f$ in $J$ and $x_n\in J$ for any $n\geq 1$.

Answer 2

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

$$L^4-5L^2+1=0$$

$$L_1^2=\frac{5-\sqrt{21}}{2}\approx 0.7$$

$$L_2^2=\frac{5+\sqrt{21}}{2}\approx 4.7$$

Answer 3

Hint:

This is not a detailed answer, but this plot says all.