Determine all integers $ n>1$ for which the inequality \[ x_1^2\plus{}x_2^2\plus{}\ldots\plus{}x_n^2\ge(x_1\plus{}x_2\plus{}\ldots\plus{}x_{n\minus{}1})x_n\] holds for all real $ x_1,x_2,\ldots,x_n$.
1. We start with the given inequality:
   \[
   x_1^2 + x_2^2 + \ldots + x_n^2 \ge (x_1 + x_2 + \ldots + x_{n-1})x_n
   \]
   for all real \( x_1, x_2, \ldots, x_n \).

2. To simplify, we rewrite the inequality by isolating \( x_n \):
   \[
   x_1^2 + x_2^2 + \ldots + x_n^2 - (x_1 + x_2 + \ldots + x_{n-1})x_n \ge 0
   \]

3. We complete the square for each term involving \( x_n \):
   \[
   x_1^2 + x_2^2 + \ldots + x_{n-1}^2 + x_n^2 - (x_1 + x_2 + \ldots + x_{n-1})x_n
   \]
   \[
   = \sum_{i=1}^{n-1} \left( x_i^2 - x_i x_n \right) + x_n^2
   \]

4. We complete the square for each \( x_i \):
   \[
   x_i^2 - x_i x_n = \left( x_i - \frac{x_n}{2} \right)^2 - \frac{x_n^2}{4}
   \]
   Therefore, the inequality becomes:
   \[
   \sum_{i=1}^{n-1} \left( \left( x_i - \frac{x_n}{2} \right)^2 - \frac{x_n^2}{4} \right) + x_n^2 \ge 0
   \]

5. Simplifying the sum:
   \[
   \sum_{i=1}^{n-1} \left( x_i - \frac{x_n}{2} \right)^2 - \frac{(n-1)x_n^2}{4} + x_n^2 \ge 0
   \]
   \[
   \sum_{i=1}^{n-1} \left( x_i - \frac{x_n}{2} \right)^2 + \left( 1 - \frac{n-1}{4} \right)x_n^2 \ge 0
   \]

6. For the inequality to hold for all \( x_1, x_2, \ldots, x_n \), the coefficient of \( x_n^2 \) must be non-negative:
   \[
   1 - \frac{n-1}{4} \ge 0
   \]
   \[
   4 - (n-1) \ge 0
   \]
   \[
   5 - n \ge 0
   \]
   \[
   n \le 5
   \]

7. We need to check if the inequality holds for \( n = 2, 3, 4, 5 \):
   - For \( n = 2 \):
     \[
     x_1^2 + x_2^2 \ge x_1 x_2
     \]
     This is true by the AM-GM inequality.
   - For \( n = 3 \):
     \[
     x_1^2 + x_2^2 + x_3^2 \ge (x_1 + x_2)x_3
     \]
     This is true by the Cauchy-Schwarz inequality.
   - For \( n = 4 \):
     \[
     x_1^2 + x_2^2 + x_3^2 + x_4^2 \ge (x_1 + x_2 + x_3)x_4
     \]
     This is true by the Cauchy-Schwarz inequality.
   - For \( n = 5 \):
     \[
     x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 \ge (x_1 + x_2 + x_3 + x_4)x_5
     \]
     This is true by the Cauchy-Schwarz inequality.

8. Therefore, the inequality holds for \( n \in \{2, 3, 4, 5\} \).

The final answer is \( \boxed{ n \in \{2, 3, 4, 5\} } \)