{
    "problem": "What are all values of $p$ such that for every $q>0$, we have   $$\\frac{3(pq^2+p^2q+3q^2+3pq)}{p+q}>2p^2q?$$ Express your answer in interval notation in decimal form.",
    "level": "Level 5",
    "type": "Algebra",
    "solution": "First we'll simplify that complicated expression. We attempt to factor the numerator of the left side: \\begin{align*}\npq^2+p^2q+3q^2+3pq &= q(pq + p^2 + 3q + 3p) \\\\\n&= q[ p(q+p) + 3(q+p) ] \\\\\n&= q(p+3)(q+p).\n\\end{align*}Substituting this in for the numerator in our inequality gives $$\\frac{3q(p+3)(p+q)}{p+q}>2p^2q.$$We note that left hand side has $p+q$ in both the numerator and denominator.  We can only cancel these terms if $p+q \\neq 0.$  Since we're looking for values of $p$ such that the inequality is true for all $q > 0,$ we need $p \\geq 0$ so that $p + q \\neq 0.$\n\nAlso because this must be true for every $q>0$, we can cancel the $q$'s on both sides. This gives  \\begin{align*}\n3(p+3)&>2p^2\\Rightarrow\\\\\n3p+9&>2p^2 \\Rightarrow\\\\\n0&>2p^2-3p-9.\n\\end{align*}Now we must solve this quadratic inequality. We can factor the quadratic as $2p^2-3p-9=(2p+3)(p-3)$. The roots are $p=3$ and $p=-1.5$. Since a graph of this parabola would open upwards, we know that the value of $2p^2 - 3p - 9$ is negative between the roots, so the solution to our inequality is $-1.5<p<3.$  But we still need $0 \\leq p,$ so in interval notation the answer is $\\boxed{[0,3)}$."
}