Solution:

At a = 0, b = c = d = t > 0 the inequality has the form: 9t2 > Kt2. Therefore, K ≤ 9. Now let us prove that K = 9 satisfies the inequality:
(a + b + c + d)2 > 9bc
(1)

Indeed,

                        b + c   3
(since d > b,d > c). By squaring both sides, we get
                   9 .
Finally, since (b + c)2 ≥ 4bc we get the desired inequality (1).

Thus, the maximal possible value of K is 9.