Upper bound for the number of integral points in a convex set Let $K \subset \mathbb{R}^3$ be a bounded convex set such that the points with integer
coordinates in $K$ are not all coplanar. Is it true that $|K \cap \mathbb{Z}^3| \leq 6{\rm Vol}(K) + 3$?
 A: It suffices to prove the bound for (non-planar) polytopes $K$ with integer vertices. Let $v$ be a vertex of $K$. By triangulating faces of $K$ which does not contain $v$ and considering the corresponding simplexes with vertex $v$ based at these triangles one can obtain a decomposition of $K$ to simplexes with integer vertices.For such simplexes one can easily prove that the number of lattice points inside and on the faces except the vertices is at most six times the volume of the simplex minus 1: 
Proof. By a well known result the number of lattice points in the parallelepiped generated by independent integer vectors $v_1,v_2,v_3$:
$$\{\lambda_1 v_1+\lambda_2 v_2 + \lambda_3 v_3 : 0\leq \lambda_1,\lambda_2,\lambda_3<1 \}$$ 
is equal to  the volume of parallelepiped which is six times the volume of the simplex with vertices $0,v_1,v_2,v_3$ and the origin with other lattice points of the simplex except vertices are in this parallelepiped.
So the number of all integer points in $K$ is at most $6 \mathrm{Vol(K)}$-$F$(the number of triangles = simplexes) + 1 (for $v$) + $V$ (the number of other vertices of $K$). But one can consider the triangulation as a connected triangulation in plane. By easy induction (adding triangles one by one) $E$ (the number of edges)  is at most $2F+1$. So by Euler's formula: (we don't consider the unbounded face)
$$V+F = E+1 \leq 2F+2 \Rightarrow V \leq F +2.$$ 
Therefore the total number of integer points of $K$ is $\leq 6\mathrm{Vol}(K)+3$.
