Belarus — Final Round

If $x \ge -1$ satisfies the problem condition, then setting $a_1 = a_2 = \dots = a_n = 1$ in the given inequality:
$$\frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \dots \cdot \frac{a_n + x}{2} \le \frac{a_1 a_2 \dots a_n + x}{2} \quad (1)$$
we obtain the inequality $\left(\frac{1 + x}{2}\right)^n \le \frac{1 + x}{2}$, which yields $x \le 1$.
We prove, by induction on $n \ge 2$, that for $x \in [-1, 1]$ inequality (1) holds for all $n \in \mathbb{N}$ and all $a_1, \dots, a_n \ge 1$.
For $n=2$ inequality (1) has the form:
$$\frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \le \frac{a_1 a_2 + x}{2}, \quad (2)$$
which is equivalent to:
$$x^2 + (a_1 + a_2 - 2)x - a_1 a_2 \le 0.$$
The last inequality holds for all $x \in [-1, 1]$ and all $a_1, a_2 \ge 1$. Indeed,
$$\begin{aligned} & x^2 + (a_1 + a_2 - 2)x - a_1 a_2 \le 1 + (a_1 + a_2 - 2) \cdot 1 - a_1 a_2 \\ & = a_1 + a_2 - a_1 a_2 - 1 = -(a_1 - 1)(a_2 - 1) \le 0. \end{aligned}$$
The base of induction is proved.
If we suppose that inequality (1) holds for some $n \ge 2$, then from (1) and (2) it follows that:
$$\begin{aligned} & \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \dots \cdot \frac{a_n + x}{2} \cdot \frac{a_{n+1} + x}{2} \le \frac{a_1 a_2 \dots a_n + x}{2} \cdot \frac{a_{n+1} + x}{2} \le \\ & \le \frac{a_1 a_2 \dots a_n a_{n+1} + x}{2}, \end{aligned} \quad (3)$$
as required.