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!

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


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