Question 49
Let $a$, $b$, $c$ and $d$ be positive real numbers such that $abcd = \frac{1}{4}$. Prove that
$$
\left(16ac + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{ac}\right) \left(bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd}\right) \ge \frac{81}{4}.
$$
When does the equality hold?
$$
\left(16ac + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{ac}\right) \left(bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd}\right) \ge \frac{81}{4}.
$$
When does the equality hold?