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inequality

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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.

Aug 15, 2023

#1
+121
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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:

$p \in (-\infty, 0).$

Aug 16, 2023