  We prove that every homogeneous convex polyhedron with only one unstable
equilibrium (known as a mono-unstable convex polyhedron) has at least $7$
vertices. Although it has been long known that no mono-unstable tetrahedra
exist, and mono-unstable polyhedra with as few as $18$ vertices and faces have
been constructed, this is the first nontrivial lower bound on the number of
vertices for a mono-unstable polyhedron.
  There are two main ingredients in the proof. We first establish two types of
relationships, both expressible as (non-convex) quadratic inequalities, that
the coordinates of the vertices of a mono-unstable convex polyhedron must
satisfy, taking into account the combinatorial structure of the polyhedron.
Then we use numerical semidefinite optimization algorithms to compute easily
and independently verifiable, rigorous certificates that the resulting systems
of quadratic inequalities (5943 in total) are indeed inconsistent in each case.
