Find values of $p$ such that, for all $q>0$, $\frac{3(pq^2+p^2q+3q^2+3pq)}{p+q}>2p^2q$

Question

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.

I have no idea how to start. Any help would be great.

Answer 1

Factor the numerator so that $(p+q)$ cancels: $$\frac{3(pq^2+p^2q + 3q^2 + 3pq)}{p+q} = \frac{3(pq(p+q) + 3q(p+q))}{p+q} = 3pq + 9q.$$ Now, since $q > 0$, cancel a $q$ on both sides, and you have a quadratic inequality $$-2p^2+3p+9>0.$$ Now I leave it to you...

Answer 2

Fix $q > 0$.

For now, assume $p\ne -q\,$; we'll address that issue later. \begin{align*} \text{Then}\;\;&\frac{3(pq^2+p^2q+3q^2+3pq)}{p+q} > 2p^2q\\[4pt] \iff\;&\frac{3q(pq+p^2+3q+3p)}{p+q} > 2p^2q\\[4pt] \iff\;&\frac{3(pq+p^2+3q+3p)}{p+q} > 2p^2&&\text{[since $q > 0$]}\\[4pt] \iff\;&\frac{3\bigl(p(p+q)+3(p+q)\bigr)}{p+q} > 2p^2\\[4pt] \iff\;&\frac{3(p+q)(p+3)}{p+q} > 2p^2\\[4pt] \iff\;&3(p+3) > 2p^2&&\text{[since $p+q \ne 0$]}\\[4pt] \iff\;&2p^2-3p-9 < 0\\[4pt] \iff\;&(p-3)(2p+3) < 0\\[4pt] \iff\;&-{\small{\frac{3}{2}}} < p < 3\\[4pt] \end{align*} But the inequality must hold for all $q > 0$, so now let's worry about the possibility that $p= -q$.

If $p< 0$, the inequality fails for the case $q=-p$.

It follows that the interval of values for $p$ for which the inequality holds for all $q > 0$ is the interval $[0,3)$.