make a change of variable. let $y_n = x_n - 1.$ then $$y_{n+1} = \sqrt y_n,\quad y_1 \ge 1.$$
we have $$1 \le y_1 \implies 1 \le y_1 \le y_1^2=y_2 $$ and by induction, we have $$y_n \le y_{n+1}, n \ge 1. $$
it can also be seen from the fact that $\sqrt x$ has a unique fixed point $x = 1$ and from its graph.
Here's a more formal (and shorter) proof:
Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$, which we know exists, and let $L$ be its limit. Then, given any $\varepsilon > 0$, there is some $K \in \mathbb{N}$ such that for all $k \geq K$, $|x_{n_k} - L| < \varepsilon$ (this is the definition of the limit).
Now, for any $n \geq n_K$, we will show than $|x_n - L| < \varepsilon$, which will complete the proof. For this purpose, suppose that this is not true. Then there is some $n \geq n_K$ such that $|x_n - L| \geq \varepsilon > |x_{n_K} - L|$.
First, note that if $x_1 \leq L$ then we must have $x_i \leq L$ for all $i$: if not, then we have some $i$, such that $x_1 \leq L < x_i$, but then for all $m > i$, $x_m \geq x_i > L$, so $|x_m - L| \geq |x_i - L| > 0$, so in particular, $(x_{n_k})\not\to L$, a contradiction. Symmetrically, if $x_1 \geq L$, then $x_i \geq L$ for all $i$.
Now, we have a problem: we have either $x_n \geq L + \varepsilon > x_{n_K} \geq L$ or $x_n\leq L - \varepsilon < x_{n_K} \leq L$, but then monotonicity gives us either $x_m \geq L + \varepsilon$ or $x_m \leq L - \varepsilon$ for all $m \geq n$, in particular for all but finitely many $n_k$, so $|x_{n_k} - L| \geq \varepsilon$ for all but finitely many $n_k$, so $(x_{n_k})\not\to L$, a contradiction.
Thus, we must have $|x_n - L| <\varepsilon$ for all $n \geq n_K$, hence $(x_n)\to L$.
Best Answer
First assume that $x_1\le \sqrt a$. Then $x_n\le \sqrt a$ for $n\ge 2$. In fact,
\begin{eqnarray} x_{n+1}-\sqrt a=\dfrac{x_n(x_n^2+3a)}{3x_n^2+a}-\sqrt a=\dfrac{(x_n-\sqrt a)^3}{3x_n^2+a} \le0 \end{eqnarray} by induction. Note $$ \frac{x_{n+1}}{x_n}-1=\dfrac{x_n^2+3a}{3x_n^2+a}-1=\frac{2(a-x_n^2)}{3x_n^2+a}\ge0$$ and hence $\{x_n\}$ is bounded and increasing. So $\{x_n\}$ converges. You can treat $x_1>\sqrt a$ similarly.