Let $p$ and $q$ be distinct positive real numbers. (a) Prove by contradiction that \(\frac{p^2}{

Question

Let $p$ and $q$ be distinct positive real numbers. (a) Prove by contradiction that $\frac{p^2}{q}+\frac{q^2}{p}>p+q$. (b) Give a counterexample to show that this inequality need not hold if $p$ and $q$ are nonzero but not both positive.

Accepted Answer

$\frac{p^2}{q}+\frac{q^2}{p}>p+q$ for unequal positive $p,q$; $p=1,\ q=-1$ gives equality.

Cubic quotient inequality

Cubic quotient inequality

Solution:

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