prove that inequality holds for all reals if and only if $|k| \leq 2n$

Prove that $$x_1^2+x_2^2+\ldots+x_{2n}^2+k \cdot x_1 \cdot x_2 \cdot \ldots \cdot x_{2n}\geq 0$$ for all reals $$x_1, x_2, \ldots, x_{2n}$$ if and only if $$|k| \leq 2n$$, where $$n$$ and $$k$$ integers, $$n > 0$$.

Ideas : Let's choose firstly $$|k| \leq 2n$$ and let's discuss only $$0 \leq k \leq 2n$$, the other case being similar. We shall use the inequality: $$x_1^2+x_2^2+\ldots+x_{2n}^2 \geq \sqrt[2n]{x_1^2 x_2^2 \dots x_n^2} = \sqrt[n]{x_1x_2 \dots x_n}$$

But, however, other noting $$\sqrt{x_1x_2 \dots x_n} = p$$ and wiring an equation in $$p, n$$ and $$k$$, I can't advance any further. Thank you!

• I do not think the claim is correct. $x_1=-1, x_2=2,x_3=3,x_4=4, k=3$ Nov 28, 2022 at 13:24