tchebytchev wrote: Find all integers $n \geq 1$ such that for all $x_1,...,x_n \in \mathbb{R}$ the following inequality is satisfied $$\left(\frac{x_1^n+...+x_n^n}{n}-x_1....x_n\right)\left(x_1+...+x_n\right) \geq 0$$ It's obvious that $n$ should be odd and $n=3$ and $n=1$ are valid. For all $n\geq5$ there is a counter-example: $x_1+...+x_{n-1}<0$, $x_n=0$, $x_1>0$ and $x_2$,...,$x_{n-1}$ are negatives.

$x_2=...=x_n=0 \to x_1^{n+1} \geq 0 \to n=2k+1$ For $n=1,3$ it is true. For $n>3$ set $x_1=2,x_2=3,x_3=-4,x_4=...x_n=0$ $2^n+3^n-4^n<2*3^n-4^n<(\frac{4}{3})^n 3^n-4^n<0$ And so $(2^n+3^n-4^n)(2+3-4)<0$

Yes that is true. The only tricky case is $n=3$ but easy once we know how to factorize $x^3+y^3+z^3-3xyz$