AoPS - Problem

Set $a_i=1$ $\frac{(1+x)^n}{2^n}\leq \frac{1+x}{2}$ $1+x\leq 2$ => $x\leq 1$ , so $x^2\leq 1$ Let $a_1\geq 1,a_2 \geq 1$ $\frac{a_1a_2+x}{2}-\frac{a_1+x}{2}*\frac{a_2+x}{2}=\frac{2a_1a_2+2x-a_1a_2-a_1x-a_2x-x^2}{4}=\frac{(a_1-x)(a_2-x)+2x-2x^2}{4}\geq$ $\geq \frac{(1-x)^2+2x-2x^2}{4}=\frac{1-x^2}{4}\geq 0$ It means that for $x^2\leq 1$ and $a_1\geq 1,a_2 \geq 1$ next inequality is true: $\frac{a_1+x}{2}*\frac{a_2+x}{2}\leq \frac{a_1a_2+x}{2}$ By induction it is easy to prove that than for $n\geq 2$ and $x^2\leq 1$: $$\frac{a_1+x}{2}*\frac{a_2+x}{2}*...*\frac{a_n+x}{2}\leq\frac{a_1a_2...a_n+x}{2}$$ So answer is $x \in [-1,1]$