Let real numbers $x_1, x_2, x_3, x_4, x_5, x_6$ satisfy $x_1+x_2+x_3+x_4+x_5+x_6=0, $ and $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2=6.$ Prove $x_1x_2x_3x_4x_5x_6\leq\frac{1}{2}.$

I am trying to figure out the question but I am currently stuck. I tried doing RMS-AM-GM inequality to try and prove this, but I made no progress. Can somebody help me? Thanks.

  • 1
    $\begingroup$ It probably helps to observe that WLOG exactly two of the $x_i$ are negative. Note also that the first constraint is important in order to bound the tuple away from $(1,1,1,1,-1,1)$. $\endgroup$ – Erick Wong May 25 '19 at 2:12

As Erick Wong says, we can WLOG that exactly two of the $x_i$ are negative, say $x_5,x_6$.

Let $S=x_1+x_2+x_3+x_4$. Then by RMS-AM, $$\sum\limits_{i=1}^4 x_i^2\geq \frac{S^2}{4}$$ and similarly $$x_5^2+x_6^2\geq \frac{S^2}{2}$$ so $S\leq 2\sqrt{2}$.

Since $\sqrt[4]{x_1x_2x_3x_4}\leq\frac{S}{4}$ and $\sqrt{x_5x_6}\leq\frac{S}{2}$, we know that $$\prod x_i\le \frac{S^6}{4^4\cdot2^2}\leq \frac{2^9}{2^{10}}=\frac{1}{2}$$ as required.


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