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.

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 $a_n \geq 2$:
$$ \color{blue}{a_{n+1}-2} = \frac{3a_n}{4}+\frac{1}{a_n} - \left(\frac{3\cdot 2}{4}+\frac{1}{2} \right)= \frac{3}{4}\left( a_n - 2 \right) - \frac{a_n - 2}{2a_n}$$
$$ = \left(\frac{3}{4} - \frac{1}{2a_n} \right)(a_n -2) \stackrel{\color{blue}{a_n \geq 2}}{\geq} \frac{1}{2}(a_n -2) \color{blue}{\geq 0}$$
