We fix a natural number $n \geqslant 2$. Determine all real numbers $x \geqslant-1$ such that for all real numbers $a_{1}, \ldots, a_{n} \geqslant 1$ we have

$$
\frac{a_{1}+x}{2} \times \cdots \times \frac{a_{n}+x}{2} \leqslant \frac{a_{1} a_{2} \cdots a_{n}+x}{2} .
$$
By setting $a_{i}=1$ for all $i$, we obtain the condition $\left(\frac{1+x}{2}\right)^{n} \leqslant \frac{1+x}{2}$. Let $y=\frac{1+x}{2}$. By hypothesis, $y \geqslant 0$. Since $y^{n} \leqslant y$, we have $y^{n-1} \leqslant 1$ and thus $y \leqslant 1$, which implies $x \leqslant 1$.
Conversely, let's show that any $x \in[-1,1]$ works. We proceed by induction on $n$. For $n=1$ the inequality is obvious. Suppose it is true at rank $n-1$. The left-hand side of the inequality is then less than or equal to $\frac{a+x}{2} \times \frac{b+x}{2}$, where $a=a_{1} a_{2} \cdots a_{n-1}$ and $b=a_{n}$. It then suffices to show that $\left(\frac{a+x}{2}\right)\left(\frac{b+x}{2}\right) \leqslant\left(\frac{a b+x}{2}\right)$. By expanding, the inequality becomes $(a+b-2) x+x^{2} \leqslant a b$. Since $a+b-2 \geqslant 0$, and since $x \leqslant 1$ and $x^{2} \leqslant 1$, the left-hand side of the inequality is bounded by $(a+b-2)+1$, so it suffices to show that $(a+b-2)+1 \leqslant a b$, or equivalently that $0 \leqslant a b-a-b+1$. Now, we have $a b-a-b+1=(a-1)(b-1)$. Since $a \geqslant 1$ and $b \geqslant 1$, the inequality $(a-1)(b-1) \geqslant 0$ is indeed satisfied.