Find all real numbers $p$ that have the following property: For all $q>0$, the inequality

$$\frac{3(pq^2+p^2q)}{p + q} +3q^2+3pq >2p^2q$$

holds.

Express your answer in interval notation.

Guest Aug 15, 2023

#1**0 **

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:

\[p \in (-\infty, 0).\]

SpectraSynth Aug 16, 2023