Question:
Suppose $a$ and $b$ are positive real numbers with $a > b$ and $ab = 8.$  Find the minimum value of $\frac{a^2 + b^2}{a - b}.$

Answer:
We can write
\[\frac{a^2 + b^2}{a - b} = \frac{a^2 + b^2 - 2ab + 16}{a - b} = \frac{(a - b)^2 + 16}{a - b} = a - b + \frac{16}{a - b}.\]By AM-GM,
\[a - b + \frac{16}{a - b} \ge 2 \sqrt{(a - b) \cdot \frac{16}{a - b}} = 8.\]Equality occurs when $a - b = 4$ and $ab = 8.$  We can solve these equations to find $a = 2 \sqrt{3} + 2$ and $b = 2 \sqrt{3} - 2.$  Thus, the minimum value is $\boxed{8}.$