Vietnamese Mathematical Olympiad

a) It is easy to see that $x_n > 0$ for all $n \in \mathbb{N}^*$. For every positive integer $n$, we have
$$\begin{aligned} |x_{n+1} - 1| &= |\sqrt{x_n + 8} - 3 + 2 - \sqrt{x_n + 3}| \\ &= |(x_n - 1)\left(\frac{1}{\sqrt{x_n + 8} + 3} - \frac{1}{\sqrt{x_n + 3} + 2}\right)| \\ &\le |x_n - 1|\left(\frac{1}{\sqrt{x_n + 8} + 3} + \frac{1}{\sqrt{x_n + 3} + 2}\right) \\ &\le |x_n - 1|\left(\frac{1}{3} + \frac{1}{2}\right) \\ &= \frac{5}{6}|x_n - 1|. \end{aligned}$$
Therefore,
$$|x_n - 1| \le \frac{5}{6}|x_{n-1} - 1| \le \dots \le \left(\frac{5}{6}\right)^{n-1} |x_1 - 1| = \left(\frac{5}{6}\right)^n, \forall n \in \mathbb{N}^*.$$
Note that $\lim \left(\frac{5}{6}\right)^n = 0$, we obtain $\lim_{n \to \infty} x_n = 1$.

b) Consider the function
$$f(x) = \sqrt{x+8} - \sqrt{x+3} = \frac{5}{\sqrt{x+8} + \sqrt{x+3}},$$
with $x > 0$, we see that $f(x)$ is a continuous and decreasing function on $(0, +\infty)$. Because $x_1 > 1$ then $x_2 = f(x_1) < f(1) = 1$, and $x_3 = f(x_2) > f(1) = 1, \dots$. In general, we can prove $x_{2k} < 1 < x_{2k-1}$ for all positive integers $k$.

Now, consider the function $g(x) = x + f(x) = x + \sqrt{x+8} - \sqrt{x+3}$ with $x > 0$, we get $g(x)$ is a continuous function and
$$g'(x) = 1 + \frac{1}{2\sqrt{x+8}} - \frac{1}{2\sqrt{x+3}} > 1 - \frac{1}{2\sqrt{3}} > 0, \forall x > 0$$
so $g(x)$ is an increasing function on $(0, \infty)$. From here, we have the following claims
* If $x > 1$ then $g(x) > g(1) = 2$.
* If $0 < x < 1$ then $g(x) < g(1) = 2$.
Hence,
$$x_{2k-1} + x_{2k} > 2 > x_{2k} + x_{2k+1}, \forall k \in \mathbb{N}^*.$$
Now, we will prove the given inequality. Consider two cases:
* Case 1: $n = 2k$ ($k \in \mathbb{N}^*$). It is easy to check that $2 < x_1 + x_2 < 3$ so the given inequality is true when $k = 1$. Assume that $k > 1$, we have
$$x_1 + x_2 + \dots + x_n = (x_1 + x_2) + (x_3 + x_4) + \dots + (x_{2k-1} + x_{2k}) \\ > 2 + 2 + \dots + 2 = 2k$$
and
$$x_1 + x_2 + \dots + x_n = x_1 + (x_2 + x_3) + \dots + (x_{2k-2} + x_{2k-1}) + x_{2k} \\ < 2 + 2 + \dots + 2 + 1 = 2k + 1.$$
* Case 2: $n = 2k - 1$ ($k \in \mathbb{N}^*$). Clearly, the given inequality is true when $k = 1$. Suppose that $k > 1$, we have
$$x_1 + x_2 + \dots + x_n = (x_1 + x_2) + \dots + (x_{2k-3} + x_{2k-2}) + x_{2k-1} \\ > 2 + 2 + \dots + 2 + 1 = 2k - 1$$
and
$$x_1 + x_2 + \dots + x_n = x_1 + (x_2 + x_3) + \dots + (x_{2k-2} + x_{2k-1}) \\ < 2 + 2 + \dots + 2 = 2k.$$
To summarize, we have $n \le x_1 + x_2 + \dots + x_n \le n + 1$ for all positive integers $n$. $\square$