case 1 again:

Without loss of generality, we may assump x<=y<=z=M

case 1. x < y, then both Y1= intsec(Y,(x,M]) and Z1=(Z,(z,M]) are non-empty, and (x,M] = union(Y1,Z1)
so, for any x1 in X, we can find y1 in Y1 and z1 in Z1, such that |y1-z1| < x1, and by construction, y1 >= x1, z1 >= x1, so x1,y1,z1 can form a triangle.