 Solution

Suppose the underlying graph has a cycle through $k\ge 3$ distinct vertices, and let $v_1,\dots ,v_k$ be their unit vectors from admissibility. Adjacent vertices have $(v_i,v_j)\le -1/2$, and every other pair has nonpositive inner product. ThereforeBut linear independence makes the sum nonzero, whose squared norm in a Euclidean space must be positive. This contradiction proves that the underlying graph of an admissible Coxeter graph has no cycles. Double and triple lines record edge multiplicities; they are not cycles of length two. The argument also tolerates chords, which only make the bound smaller.