Vietnam Mathematical Olympiad

For $n \ge 1$, we have
$$x_{n+1} - x_n = \left(\frac{n+3}{3(n+1)} - 1\right)x_n + \frac{2(n+3)}{3(n+1)} = \frac{2}{3(n+1)}(n+3 - nx_n). \quad (1)$$
We first prove that
$$x_n > 1 + \frac{3}{n} \quad \forall n \ge 2. \quad (2)$$
The proof proceeds by induction on $n$. For $n=2$ we have
$$x_2 = \frac{4}{6}(3+2) = \frac{10}{3} > 1 + \frac{3}{2}.$$
Suppose that $x_k > 1 + \frac{3}{k}$ for some $k \ge 2$, then we have
$$x_{k+1} = \frac{k+3}{3(k+1)}(x_k + 2) > \frac{k+3}{3(k+1)}\left(1 + \frac{3}{k} + 2\right) = 1 + \frac{3}{k+1}.$$
Thus, (2) is proved. From (1) and (2), we have
$$x_n > x_{n+1} > 1, \quad \forall n \ge 2.$$
Therefore, $(x_n)$ is a decreasing sequence for $n \ge 2$ and is bounded by 1. This implies that $(x_n)$ has a finite limit. Now, we can easily find this limit by solving the equation
$$x = \frac{1}{3}(x+2),$$
which implies that the limit is 1.