# Proving an inequality given conditions.

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.

• 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 $(1,1,1,1,-1,1)$. – 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.