# Proof of the the inequality

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 $$\sigma_4=a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4 + a_2a_3a_4 > 0?$$

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

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.