We first give a preliminary estimate: for any graph with vertices and edges, we have : \operatorname{cr}(G) \geq e - 3n. To prove this, consider a diagram of which has exactly crossings. Each of these crossings can be removed by removing an edge from . Thus we can find a graph with at least edges and vertices with no crossings, and is thus a
planar graph. But from
Euler's formula we must then have , and the claim follows. (In fact we have for ). To obtain the actual crossing number inequality, we now use a
probabilistic argument. We let be a
probability parameter to be chosen later, and construct a
random subgraph of by allowing each vertex of to lie in independently with probability , and allowing an edge of to lie in if and only if its two vertices were chosen to lie in . Let and denote the number of edges, vertices and crossings of , respectively. Since is a subgraph of , the diagram of contains a diagram of . By the preliminary crossing number inequality, we have :\operatorname{cr}_H \geq e_H - 3n_H. Taking
expectations we obtain :\mathbf{E}[\operatorname{cr}_H] \geq \mathbf{E}[e_H] - 3 \mathbf{E}[n_H]. Since each of the vertices in had a probability of being in , we have . Similarly, each of the edges in has a probability of remaining in since both endpoints need to stay in , therefore . Finally, every crossing in the diagram of has a probability of remaining in , since every crossing involves four vertices. To see this consider a diagram of with crossings. We may assume that any two edges in this diagram with a common vertex are disjoint, otherwise we could interchange the intersecting parts of the two edges and reduce the crossing number by one. Thus every crossing in this diagram involves four distinct vertices of . The diagram we have got may be not optimal in terms of number of crossings, but it obviously has at least crossings. Therefore, : p^4 \operatorname{cr}(G) \geq \mathbf{E} [ \operatorname{cr}_H ] and we have : p^4 \operatorname{cr}(G) \geq p^2 e - 3 p n. Now if we set (since we assumed that ), we obtain after some algebra : \operatorname{cr}(G) \geq \frac{e^3}{64 n^2}. A slight refinement of this argument allows one to replace by for . ==Variations==