<s>[INST] Proving a sequence is bounded from below Let $$a_1=3\quad , \quad a_{n+1}=\frac{3a_n}{4}+\frac{1}{a_n}$$
Show that $a_n$ converges.
I know that I need to prove that $a_n$ is monotonic and bounded. 
I've assumed that the limit exists, made the calculation and get that $L=2$, hence claiming two claims:


*

*$a_n\ge 2$ (I know that 0 is a simpler bound)

*$a_n$ is monotonically decreasing.


Now, if I can prove the first claim. then I can write $$a_{n+1}=\frac{3a_n}{4}+\frac{1}{a_n}=\frac{3a_n}{4}+\frac{a_n}{a_n^2}\le \frac{3a_n}{4}+\frac{a_n}{4}=a_n$$ which proves the second claim.
My only problem is proving the first one. If it is not the right direction, please hint me.
 [/INST] To show boundedness from below, you may use AM-GM:
$$\frac{3a_n}{4}+\frac{1}{a_n}\geq 2 \sqrt{\frac{3a_n}{4}\cdot \frac{1}{a_n}}= \sqrt{3}$$
For convergence you may proceed as follows:


*

*$f(x) = \frac{3x}{4} + \frac{1}{x}$ has a fixpoint for $x^{\star} = 2$

*$f'(x) = \frac{3}{4}-\frac{1}{x^2}$

*A quick calculation shows that $|f'(x)| < 1$ for $x > \frac{2}{\sqrt{7}} \Rightarrow |f'(x)| \leq \color{blue}{q} < 1$ for $x \geq \sqrt{3}$
So, for any starting value $a_0 > 0$ you get 
$$|2-a_{n+1}| = |f'(\xi_n)||2-a_n| < \color{blue}{q} |2-a_n| < \color{blue}{q^n}|2-a_1|\stackrel{n\to \infty}{\longrightarrow}0 $$
Edit after comment:
Specifically for your question concerning $a_n \geq 2$:
$$ \color{blue}{a_{n+1}-2} = \frac{3a_n}{4}+\frac{1}{a_n} - \left(\frac{3\cdot 2}{4}+\frac{1}{2} \right)= \frac{3}{4}\left( a_n - 2 \right) - \frac{a_n - 2}{2a_n}$$
$$ = \left(\frac{3}{4} - \frac{1}{2a_n} \right)(a_n -2) \stackrel{\color{blue}{a_n \geq 2}}{\geq} \frac{1}{2}(a_n -2) \color{blue}{\geq 0}$$
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