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?

Suppose $(x_n)$ converges, say $\lim x_n = l \in \mathbb{R}$. Since the function $f : x \mapsto \frac{x^3 + 3 a x}{3 x^2 + a}$ is continuous, we have $\lim f(x_n) = f(\lim x_n) = f(l)$, but we also have $\lim f(x_n) = \lim x_{n+1} = \lim x_n = l$. Hence, we must have $l = f(l)$.
Possible limits
The equation $f(x) = x$ has three solutions : $-\sqrt{a}$, $0$ and $\sqrt{a}$.
In particular, the answer to the problem is no. Take eg $x_0 = 0$, then $x_n = 0$ for all $n$ hence $\lim x_n = 0$.
limit for a given $x_0$
For this, we need to solve the inequation $f(x)>x$, hence to establish the variation chart of the function $g$ defined by $g(x) = f(x) - x$. 
You'll get that :
$f(x)> x \Longleftrightarrow x \in ]- \infty;-\sqrt{a}[\cup ]0;\sqrt{a}[$  and
$f(x)< x \Longleftrightarrow x \in ]-\sqrt{a};0[\cup ]\sqrt{a};+\infty[$
To sum it up :


*

*If $x_0 < -\sqrt{a}$, then for all $n$, $x_n < f(x_n) < -\sqrt{a}$ hence the sequence $(x_n)$ converges since it is increasing and majored. We get $\lim x_n = -\sqrt{a}$ in this case.

*If $-\sqrt{a} < x_0 < 0$, then for all $n$, $ -\sqrt{a} < f(x_n) < x_n$. With a similar argument, we get $\lim x_n = -\sqrt{a}$ in this case.

*Using the symetry ($f$ is odd), one gets that $\lim x_n = \sqrt{a}$ whenever $x_0>0$.

