FINAL ROUND

If $x\ge -1$ satisfies the problem condition, then setting $a_1=a_2=\cdots =a_n=1$ in the given inequality: $$\frac{a_1+x}{2}\cdot \frac{a_2+x}{2}\cdot \cdots \cdot \frac{a_n+x}{2}\le \frac{a_1{a}_2\dots a_n+x}{2}(1)$$ 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: $$\frac{a_1+x}{2}\cdot \frac{a_2+x}{2}\le \frac{a_1{a}_2+x}{2},(2)$$ 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, $$\begin{array}{rl}& x^2+(a_1+a_2-2)x-a_1{a}_2\le 1+(a_1+a_2-2)\cdot 1-a_1{a}_2\\ & =a_1+a_2-a_1{a}_2-1=-(a_1-1)(a_2-1)\le 0.\end{array}$$ 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: $$\begin{array}{rl}& \frac{a_1+x}{2}\cdot \frac{a_2+x}{2}\cdot \cdots \cdot \frac{a_n+x}{2}\cdot \frac{a_{n+1}+x}{2}\le \frac{a_1{a}_2\dots a_n+x}{2}\cdot \frac{a_{n+1}+x}{2}\le \\ & \le \frac{a_1{a}_2\dots a_n{a}_{n+1}+x}{2},\end{array}(3)$$ as required.