Actually it's the symmetry property of metrics (d(p,q)=d(q,p)), not the triangle inequality, that guarantees the absence of a Condorcet cycle in the triangle case:
Suppose that A, B, and C are the only candidates, and that voters are concentrated very near their favorites. Assume that preferences are determined by distances, i.e. a voter prefers a nearer candidate over a more distant one. Without loss in generality, assume that d(A,B) is the greatest of the three distances. Then we have the preference profile: x: A>C>B y:B>C>A z: C >???? Suppose, by way of contradiction, that we have a Condorcet cycle. The without loss in generality suppose that the cycle is A beats B beats C beats A. The first step in this beat path ( A beats B) implies (x+z)>y, while the second step (B beats C) implies y>(x+z). We got into this contradiction by assuming the existence of a cycle. So that assumption is untenable. We have made tacit use of the symmetry property of distance by assuming that the longest side of the triangle was agreed upon by both A and B, i.e. A thought B was the most distant candidate, and B considered A to be the most distant candidate. To show that this property is essential, suppose that we use taxicab "distance" is a grid of one way streets, and that A, B, and C are located three of the corners [say (0,0), (1,0), and (0,1)] of a block about which the one way streets have a counterclockwise orientation. Then (by taxi) it is quicker to go from A to B than from A to C, from B to C than from B to A, and from C to A than from C to B. The preference profile has to be of the form... x: A>B>C y: B>C>A z: C>A>B which leads to a Condorcet cycle whenever no candidate has a majority. ---- Election-Methods mailing list - see http://electorama.com/em for list info