Find all real $x\geq-1$ such that for all $a_1,...,a_n\geq1$, where $n\geq2$  the following inequality holds $$\frac{a_1+x}{2}*\frac{a_2+x}{2}*...*\frac{a_n+x}{2}\leq\frac{a_1a_2...a_n+x}{2}$$
1. **Initial Setup and Simplification:**
   We start by setting \( a_i = 1 \) for all \( i \). The given inequality becomes:
   \[
   \frac{(1+x)^n}{2^n} \leq \frac{1+x}{2}
   \]
   Simplifying this, we get:
   \[
   \left(\frac{1+x}{2}\right)^n \leq \frac{1+x}{2}
   \]
   Since \( \frac{1+x}{2} \geq 0 \), we can divide both sides by \( \frac{1+x}{2} \) (assuming \( 1+x \neq 0 \)):
   \[
   \left(\frac{1+x}{2}\right)^{n-1} \leq 1
   \]
   For \( n \geq 2 \), this implies:
   \[
   \frac{1+x}{2} \leq 1
   \]
   Therefore:
   \[
   1+x \leq 2 \implies x \leq 1
   \]

2. **Verification for \( x \geq -1 \):**
   Next, we need to verify if \( x \geq -1 \) satisfies the inequality for all \( a_i \geq 1 \). Consider the case \( n = 2 \):
   \[
   \frac{a_1a_2 + x}{2} - \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2}
   \]
   Simplifying the right-hand side:
   \[
   \frac{a_1a_2 + x}{2} - \frac{(a_1 + x)(a_2 + x)}{4}
   \]
   Expanding and simplifying:
   \[
   \frac{2a_1a_2 + 2x - (a_1a_2 + a_1x + a_2x + x^2)}{4} = \frac{a_1a_2 - a_1x - a_2x + x - x^2}{4}
   \]
   Factoring out common terms:
   \[
   \frac{(a_1 - x)(a_2 - x) + 2x - 2x^2}{4}
   \]
   Since \( a_1, a_2 \geq 1 \), we have:
   \[
   \frac{(1 - x)^2 + 2x - 2x^2}{4} = \frac{1 - x^2}{4} \geq 0
   \]
   This holds if \( x^2 \leq 1 \), which implies \( -1 \leq x \leq 1 \).

3. **Induction for General \( n \):**
   We use induction to prove that for \( n \geq 2 \) and \( x^2 \leq 1 \):
   \[
   \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \ldots \cdot \frac{a_n + x}{2} \leq \frac{a_1a_2 \ldots a_n + x}{2}
   \]
   **Base Case:**
   For \( n = 2 \), we have already shown that:
   \[
   \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \leq \frac{a_1a_2 + x}{2}
   \]
   **Inductive Step:**
   Assume the inequality holds for \( n = k \):
   \[
   \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \ldots \cdot \frac{a_k + x}{2} \leq \frac{a_1a_2 \ldots a_k + x}{2}
   \]
   We need to show it holds for \( n = k+1 \):
   \[
   \frac{a_1 + x}{2} \cdot \frac{a_2 + x}{2} \cdot \ldots \cdot \frac{a_k + x}{2} \cdot \frac{a_{k+1} + x}{2} \leq \frac{a_1a_2 \ldots a_k a_{k+1} + x}{2}
   \]
   By the inductive hypothesis:
   \[
   \frac{a_1a_2 \ldots a_k + x}{2} \cdot \frac{a_{k+1} + x}{2} \leq \frac{(a_1a_2 \ldots a_k)(a_{k+1}) + x}{2}
   \]
   This completes the induction.

The final answer is \( \boxed{ x \in [-1, 1] } \).