Show that the sum of the reciprocals of three different positive integers is greater than 6 times the reciprocal of their product.

If a, b, c are the 3 positive integers

1/a + 1/b + 1/c > 6/abc

(bc+ac+ab) / abc >6/abc so

(bc+ac+ab) >6

The lowest positive integers that are different are 1,2,3 so the lowest value that (bc+ac+ab) could have is 1•2+2•3+1•3=2+6+3 = 11 therefore

1/a + 1/b + 1/c > 6/abc is true