Vietnamese Mathematical Olympiad

Firstly, by induction, one can prove that
$$a_n > 1, \quad \forall n > 1$$
because the function $u(x) = 2^{x+1} - x - 2$ is increasing on $(1, +\infty)$. It implies that
$$a_{n+1} = 3 - \frac{a_n + 2}{2^{a_n}} < 3, \quad \forall n \ge 1.$$
Next, we will prove that $(a_n)$ is an increasing sequence. Considering the function $f(x) = 3 - \frac{2+x}{2^x}$ where $1 < x < 3$. We have
$$f'(x) = \frac{\ln 4 + x \ln 2 - 1}{2^x} > 0,$$
so $f(x)$ is increasing on $(1, 3)$. Moreover $a_2 = \frac{3}{2} > a_1$ so $(a_n)$ is an increasing sequence. Note that $(a_n)$ is bounded above by $3$ so $(a_n)$ has a finite limit. Let $L \in (1, 3)$ be that limit. Letting $n$ tends to infinity, we have $L = 3 - \frac{L+2}{2^L}$. We will prove the equation
$$x = 3 - \frac{x+2}{2^x} \quad (1)$$
has a unique solution on $(1, 3)$. Indeed, consider the function $g(x) = 3 - \frac{x+2}{2^x} - x$ where $x \in (1, 3)$, we have
$$g'(x) = \frac{\ln 4 + x \ln 2 - 1}{2^x} - 1 = \frac{\ln 4 + x \ln 2 - 1 - 2^x}{2^x},$$
Setting $h(x) = \ln 4 + x \ln 2 - 1 - 2^x$, for $x \in (1, 3)$ then
$$h'(x) = \ln 2(1 - 2^x) < 0, \quad \forall x \in (1, 3),$$
so $h(x)$ is a decreasing function on $(1, 3)$. Thus, $h(x) < h(1) = \ln 8 - 3 < 0$, or $\ln 4 + x \ln 2 - 1 - 2^x < 0$ for all $x \in (1, 3)$. Thus, the function $g(x)$ is decreasing on $(1, 3)$ and the equation $g(x) = 0$ has no more than one solution.
Furthermore, $g(2) = 0$ so $x = 2$ is the unique solution of (1) which means $L = 2$ is the limit of the given sequence. $\square$