Problem - 4421
Let $a$ and $b$ be two positive real numbers satisfying $(a-b)^2=4(ab)^3$. Find the minimal value of $\frac{1}{a}+\frac{a}{b}$.
The given condition implies $$(a+b)^2 = (a-b)^2 + 4ab = 4(ab)^3+4ab$$
Then $$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\sqrt{\frac{4(ab)^3+4ab}{(ab)^2}}=2\sqrt{\frac{ab}{1}+\frac{1}{ab}}\ge \boxed{2\sqrt{2}}$$