Is there a point in 6-dimensional space satisfying these polynomial inequalities? I would like to know if there is a point $(a, b, p, q, x, y) \in [0,1]^6$ satisfying the following collection of inequalities.
$b \ge a$
$q \ge p$
$y \ge x$
$a \ge p \ge a^2$
$b \ge q \ge b^2$
$p \ge x \ge a p$
$q \ge y \ge bq$
$x \ge 2p - a$
$y \ge 2q - b$
$a^3 -3ap + 2x \ge 0$
$b^3- 3 b q +2 y \ge 0$
$(a/2+b/2)^3-3 (a/2+b/2) (p/2+q/2)+x+y < 0$
This seems like the sort of question that should be possible to answer, if not by hand, then with some help from a computer. Unfortunately, I am not an expert, so my attempts to work this out with Mathematica have not been successful. Any help would be greatly appreciated!
 A: All the conditions hold for
$$(a, b, p, q, x, y)=\left(\frac{211}{500},\frac{531}{1000},\frac{96106069}{341750000},\frac{281961}{1000000},\frac{23996819}{170875000},\frac{149721291}{1000000000}\right).$$

This result was obtained by solving the system of all the (linear) inequalities containing $x$ but not $y$ and thus getting $p\ge x\ge \max(a p,2p-a,(3a p-a^3)/2)$ (which implies $p\le a$). Then piecewise expand $\max(a p,2p-a,(3a p-a^3)/2)$; note the three corresponding cases -- call them the $x$-cases. Similarly do with the roles of $x$ and $y$ interchanged, to get the three corresponding $y$-cases. In each of the 9 combinations of the $x$- and $y$-cases, minimize the left-hand side of the last inequality in $x,y$, then in $p,q$, and finally in $a,b$. Actually, it is the combination of the $x$-case when $\max(a p,2p-a,(3a p-a^3)/2)=(3a p-a^3)/2$ and the similar $y$-case that gives the above result.

Below is the image of a piece of a Mathematica notebook providing a verification of the above statement (click on the image to enlarge it):

