Let A be a positive integer. Let D(A) be the number of solutions of xyz=A where x, y, and z are all positive integers; thus, D(6)=9. Let d(A) be the number of positive divisors of A, so d(6)=4.

Show that d(A)²-2D(A)+d(A)≥0. When does the equality hold?

(In reply to solution with proof by Daniel)

its d(A)²-2D(A)+d(A)≥0. not D(A)^2-2D(A)+d(A)>=0.

Posted by Praneeth on 2007-08-13 03:25:56