Let's analyze the given inequality step by step:
\[\frac{3(pq^2+p^2q)}{p + q} + 3q^2 + 3pq > 2p^2q.\]
We can start by simplifying the left side of the inequality:
\[\frac{3(pq^2+p^2q)}{p + q} + 3q^2 + 3pq = \frac{3pq(p + q)}{p + q} + 3q^2 + 3pq = 3pq + 3q^2 + 3pq = 6pq + 3q^2.\]
So the inequality becomes:
\[6pq + 3q^2 > 2p^2q.\]
Dividing both sides by \(3q\), where \(q > 0\):
\[2p + q > \frac{2p^2q}{3q}.\]
Simplify the right side:
\[2p + q > \frac{2p^2}{3}.\]
Now, let's analyze the cases when \(p > 0\) and \(p < 0\):
**Case 1: \(p > 0\)**
If \(p > 0\), then the inequality becomes:
\[2p + q > \frac{2p^2}{3}.\]
As \(p\) increases, the left side increases linearly, and the right side increases quadratically. For large values of \(p\), the quadratic term dominates, and the inequality is not satisfied.
**Case 2: \(p < 0\)**
If \(p < 0\), then the inequality becomes:
\[2p + q > \frac{2p^2}{3}.\]
As \(p\) becomes more negative, the left side becomes more negative, and the right side becomes more positive. This indicates that the inequality holds for all \(q > 0\) when \(p < 0\).
In conclusion, the values of \(p\) that satisfy the given inequality for all \(q > 0\) are \(p < 0\). Therefore, the solution in interval notation is:
\[p \in (-\infty, 0).\]
