4. determine all natural numbers $n$ so that for any real numbers $x_{1}, \ldots, x_{n}$ applies:

$$
\left(\frac{x_{1}^{n}+\ldots+x_{n}^{n}}{n}-x_{1} \cdot \ldots \cdot x_{n}\right)\left(x_{1}+\ldots+x_{n}\right) \geq 0
$$
Solution (Vivi and Jana): It works trivially for one. The inequality is satisfied for three. The inequality can be factorized as follows:

$$
\frac{1}{6}\left(x_{1}+x_{2}+x_{3}\right)^{2}\left(\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}\right) \geq 0
$$

It does not work for all other figures. We provide counterexamples:

- $n=2: x_{1}=0, x_{2}=-1$.
- $n>3$ : $x_{1}=x_{2}=-2, x_{3}=3, x_{4}=\ldots=x_{n}=0$. Since the second bracket is -1, we still have to show that the first bracket is positive. This is of the form

$$
\left(\frac{(-2)^{n}+(-2)^{n}+3^{n}+0^{n}+\ldots+0^{n}}{n}-(-2) \cdot(-2) \cdot 3 \cdot \ldots \cdot 0\right)=\frac{2 \cdot(-2)^{n}+3^{n}}{n}
$$

Exclusion leads to

$$
\frac{8 \cdot(-2)^{n-2}+9 \cdot 3^{n-2}}{n}
$$

This is positive for $n \geq 2$.

## Marking Scheme:

- The case $n=3$ gives 3 points.
- An upper bound is worth 1 point.
- The remaining cases are worth 3 points, with 1 point deducted for each forgotten case.