Three of the following groups of inequalities are impossible. Which group CAN have all its relations true at the same time?

Each group is a loop, so compose it right round and see what it says about X and itself. 'X ≥ Y, Y ≥ Z, Z ≥ X' composes to X ≥ Y ≥ Z ≥ X, that is X ≥ X — true of every number, and the loop simply forces X = Y = Z, so X = Y = Z = 4 satisfies it. Each of the others contains at least one strict link in a loop that otherwise points the same way, which makes the composite strict and therefore impossible: 'X > Y, Y ≥ Z, Z ≥ X' gives X > X, 'X ≥ Y, Y > Z, Z = X' gives X > X, and 'X < Y, Y ≤ Z, Z < X' gives X < X. A loop of relations is consistent precisely when every link in it is non-strict.