(A) If x > 0 and y > 0 then (x+y)n > xn+yn when n > 1 and (x+y)n< xn+yn when 0 < n < 1 So (x+y)1∕n < x1∕n+y1∕n when n > 1 Now assume x = b and y = a – b Then (b+(a–b))1∕n < b1∕n+(a–b)1∕n ⇒ a1∕n–b1∕n < (a–b)1∕n ∀ n ≥ 2 I(n) < J(n) ∀ n ≥ 2