Inequality of arithmetic and geometric means for the lattice polytopes? Let $K$,$L$,$M$ be convex lattice polytopes (so their vertices are in $\mathbb{Z}^n$) in $\mathbb{R}^n$ satisfying $K+L\subseteq M+M$ (Minkowski sum). Do we always have
$$|K\cap\mathbb{Z}^n|\cdot|L\cap\mathbb{Z}^n|\le |M\cap\mathbb{Z}^n|^2 ?$$
I looked at books/papers on Erhart polynomial and Brunn-Minkowski inequality etc., but did not find an satisfying answer. Any comment? Thanks.
 A: This inequality does not necessarily hold, at least for $n\geq 3$. It is somehow connected with the fact that there is no Pick's formula in more than two dimensions since there exists a convex lattice polytope with a large volume but containing a small number of lattice points. 
So, for instance, for $n=3$ let $M$ be the convex hull of the points $(0,0,0)$, $(1,1,0)$, $(0,1,2k)$ and $(1,0,2k)$. Then $|M\cap {\mathbb Z}^3|=4$, but $(M+M)\cap {\mathbb Z}^3\supset \{(1,1,t):0\leq t\leq 2k\}$. So $K$ and $L$ can be chosen as vertical segments containing $k$ lattice points each.
A: Actually, @Yoav's answer is more relevant than he thinks. In this paper:
MR1837217 (2002g:52011) 
Gardner, R. J.(1-WWA); Gronchi, P.(I-CNR-GA)
A Brunn-Minkowski inequality for the integer lattice. (English summary) 
Trans. Amer. Math. Soc. 353 (2001), no. 10, 3995–4024 (electronic). 
The authors prove just what they say they prove. The form of the inequality is a little different, but for example, using @Yoav's computation (which I am loath to copy and paste, he might want to undelete his answer), you get the following:
$|(K+L)/2|^2 \geq |K|(|L| - n)/n!$ in dimension $n$ (assuming $L$ is full-dimensional). They have better inequalities in $2$ dimensions, but you should just read the (very well written) paper.
A: Undeleted for the sake of a full record
Your inequality seems to follow easily from the Brunn-Minkowski inequality. Namely,
$$
|(K+L)/2|^2 \ge \left[|K/2|^{1/n}+|L/2|^{1/n}\right]^{2n}
\ge\left[4|K/2|^{1/n}|L/2|^{1/n}\right]^{n}
=|K||L|\text.
$$
