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Let $x = (x_1,x_2,x_3,x_4,x_5)$ and $y = (y_1,y_2,y_3,y_4,y_5)$ be two vectors where each $x_i,y_i\in[0,1]$ and $\sum_{i}x_i = \sum_iy_i = 1$.

In addition, assume that the following conditions are true, \begin{align} y_1x_3&\le x_1y_3\\ y_1x_4&\le x_1y_4\\ y_1x_5&\le x_1y_5\\ y_2x_4&\le x_2y_4\\ y_2x_5&\le x_2y_5\\ y_3x_4&\le x_3y_4\\ y_3x_5&\le x_3y_5\\ y_4x_5&\le x_4y_5\\ \end{align} Then, the following holds \begin{align} x_1+x_3+x_4+x_5&\le y_1+y_3+y_4+y_5\\ x_2+x_4+x_5&\le y_2+y_4+y_5\\ x_3+x_4+x_5&\le y_3+y_4+y_5\\ x_4+x_5&\le y_4+y_5\\ x_5&\le y_5 \end{align}

I have not been able to prove the above, and, as of yet, have also been unable to find a counterexample.

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    $\begingroup$ I just realized that the inequalities in my post form one big down arrow, please don't downvote! :) $\endgroup$ – jonem Jul 26 '17 at 19:18
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    $\begingroup$ you have not assigned anything for the determinants with index pairs 1,2 and 2,3 $\endgroup$ – Will Jagy Jul 26 '17 at 19:40
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Counterexample: If you set $x_1=y_2=1$, the rest of the variables have to be zero.

Now all of the top inequalities reduce to $0\le0$ and the first bottom inequality becomes $1\le0$.

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