Convergence of $x_{n+1} = \frac{(x_n)^3 + 3ax_n}{3(x_n)^2+a} $ 
Let $a \geq 0$ and let $(x_n)_n$ be a sequence such that $$x_{n+1} = \frac{(x_n)^3 + 3ax_n}{3(x_n)^2+a} $$ with $x_0 \geq 0$. Does this sequence converge to $\sqrt{a}$ for every $x_0 \geq 0$? 

If $x_0 > \sqrt{a}$, then we can easily prove that the derivative of the function $ \displaystyle x \mapsto \frac{x^3+3ax}{3x^2+a} $ is bounded above by $\frac{1}{3}$ and below by $0$, so we get that $x_n \to \sqrt{a}$ as $n \to \infty$.
However, I don't know how to approach this problem if $x_0 < \sqrt{a}$. I think that $x_n \not\to \sqrt{a}$ as $n \to \infty$ if $x_0 < \sqrt{a}$. Using the same idea as before, we can get that $x_n < \sqrt{a}, \forall n \in \mathbb{Z}_+$. Then I tried to prove that the sequence is decreasing, so it will be convergent, but with limit smaller than $\sqrt{a}$. However, I have not been able to do that. How should I proceed?

The map $\displaystyle\;\varphi: x \mapsto \frac{x^3 + 3ax}{3x^2 + a}$ has $3$ fixed points: $0, \pm \sqrt{a}$. 
Instead of the sequence $x_{n+1} = \varphi(x_n)$, one can study auxiliary sequence of the form $z_n = f(x_n)$ where $f(x)$ is a rational function over $x$
depending on the fixed points of $\varphi$.
If one is lucky, with suitable choice of $f(x)$, the recurrence sequence of $z_n$ will be significantly simpler than that of $x_n$. This will allow us to extract the asymptotic behavior of $x_n$ more easily.
In this case, if one define two auxiliary sequences $y_n, z_n$ by $$x_n = \sqrt{a} y_n\quad\text{ and }\quad z_n = \frac{1-y_n}{1+y_n} \iff y_n = \frac{1-z_n}{1+z_n}$$
One can verify $z_n$ satisfies a recurrence relation
$$z_{n+1} = z_n^3, \forall n \quad\implies\quad z_n = z_0^{3^n}$$
For any $x_0 > 0$, $|z_0| < 1$ and hence 
$$\lim_{n\to\infty} z_n = \lim_{n\to\infty} z_0^{3^n} = 0 \implies \lim_{n\to\infty} y_n = 1 \implies \lim_{n\to\infty} x_n = \sqrt{a}$$
When $x_0 = 0$, $z_0 = 1$ and hence all $z_n = 1 \implies y_n = 0 \implies x_n = 0$.
In this case, the limit is $0$ instead.
