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I have the following conditions: $$a_1<0; \quad a_2, \; a_3, \; a_4 >0$$ $$\sigma_{2, 1} = a_2a_3+a_3a_4+a_4a_2 > 0$$ $$\sigma_{2, 2} = a_1a_3+a_3a_4+a_4a_1 > 0$$ $$\sigma_{2, 3} = a_1a_2+a_2a_4+a_4a_1 > 0$$ $$\sigma_{2, 4} = a_1a_2+a_2a_3+a_3a_1 > 0$$

Is it true, that $$\sigma_4=a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4 + a_2a_3a_4 > 0?$$

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  • $\begingroup$ Are you sure $\sigma_{2, 1} = a_2a_3 + a_3a_4 + a_4a_1$ and not $a_2a_3 + a_3a_4 + a_2a_4$? $\endgroup$ – green frog Jul 3 '17 at 13:19
  • $\begingroup$ @ntntnt Your expression is obviously positive. $\endgroup$ – Michael Rozenberg Jul 3 '17 at 13:43
  • $\begingroup$ Right, it just seemed as if $\sigma_{2, i}$ was just the sum of all pairs without $a_i.$ $\endgroup$ – green frog Jul 3 '17 at 13:45
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There are an infinite number of counter-examples.

For instance, let $a_2 = a_3 = a_4 =1$. Then any $a_1$ such that -$1/2 < a_1 < -1/3$ is a counter-example.

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  • $\begingroup$ Out of curiosity how did you figure this out? $\endgroup$ – green frog Jul 3 '17 at 23:19
  • $\begingroup$ @ntntnt It's immediately obvious. $\endgroup$ – Mark L. Stone Jul 3 '17 at 23:21
  • $\begingroup$ Upon second look you are correct. $\endgroup$ – green frog Jul 3 '17 at 23:26

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