Problem 1408

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 $\boxed{ n \in \{2, 3, 4, 5\} }$