These four inequalities define a hollow (zero volume) cube. Is it possible to describe a hollow cube using a smaller system of (real scalar generalized-polynomial) inequalities?
\begin{eqnarray} x^2 & \leq & 1 \\ y^2 & \leq & 1 \\ z^2 & \leq & 1 \\ (x^2 - 1)^2(y^2 - 1)^2(z^2 - 1)^2 & \leq & 0 \end{eqnarray}
EDIT: After Rahul made me think in terms of norms, I came up with the following very simple form based on the 1-norm:
$\left ( \sqrt{x^2} + \sqrt{y^2} + \sqrt{z^2} - 1 \right ) ^2 \leq 0$
Rahul is right, the 1-isonorm and $\infty$-isonorm are similar in $\Re^2$ but not in $\Re^3$
$\left ( \left ( \sqrt{x^{2p}} + \sqrt{y^{2p}} + \sqrt{z^{2p}} \right )^{1/p} - 1 \right ) ^2 \leq 0$
approaches a cube in the limit as $p \rightarrow \infty$