For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges? $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$
For what values of $x_0$ does this sequence converge?

If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\right]=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}^2+6}{2x_{n-1}}\right]=
\frac{\lim_\limits{n \to \infty}x_{n-1}^2+6}{\lim_\limits{n \to \infty}2x_{n-1}}=
\frac{L^2+6}{2L} \implies 2L^2=L^2+6 \implies L=\sqrt{6}.$
How to determine what values of $x_0$ will give the convergence of $(x_n)$?
 A: As Professor Vector noted in the comments, you can apply AM-GM inequality for all $n\geq 1$:
$$x_n = \frac{x_{n-1}}{2}+ \frac 3{x_{n-1}} \geq 2 \sqrt{\frac{x_{n-1}}{2}\cdot \frac 3{x_{n-1}}} = \sqrt 6.$$
(To apply AM-GM, you need an observation that $x_n>0$, for all $n$, since $x_0>0$.)
Therefore, we can disregard $x_0$ from the sequence and look at $(x_n)_{n\geq 1}$ instead - convergence doesn't change, but we now have $x_n\geq \sqrt 6,\ n\geq 1$.
Now, notice that $x_n\geq \sqrt 6$ implies
$$x_{n+1} = \frac{x_n}{2}+\frac{3}{x_n} \leq \frac{x_n}{2} + \frac 3{\sqrt 6}\leq x_n.$$ (The last inequality is equivalent to $x_n\geq \sqrt 6.$)
Thus, the sequence $(x_n)_{n\geq 1}$ is decreasing and bounded from below, hence convergent. Therefore, the original sequence is convergent as well, for all values $x_0 > 0$.
A: $x_n > 0 $ for all $n$
$x_{n} \ge \sqrt{6}$ for all $n \ge 1$ by the AM-GM inequality
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}=x_{n-1}$ when $x_{n-1} = \sqrt{6}$
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}<x_{n-1}$ when $x_{n-1} > \sqrt{6}$
Basically its going decrease till $x_n = \sqrt{6}$
Therefore, it converges to $\sqrt{6}$ irrespective of the value of $x_0$
