According to the given information bcy g ax ⇒ bcy < ax ...(i) cy α bz ⇒ cy = bz ...(ii) and a2 γ bc ⇒ a2 < bc ...(iii) From (ii), we get c2y = bcz > a2z [∴ bc > a2 from (iii)] ⇒ c2y > a2y From (i), we get ax > bcy I > a2y [∴ bc > a2 from (iii)] ⇒ ax > a2y ⇒ x > ay ⇒ cx > acy ⇒ cx > abz [∴ cy = bz from (ii)] ⇒ cx ≠ abz ⇒ cx d abz.