Let $a, b, c$ and $d$ are positive real numbers so that $abcd = \frac14$. Prove that holds
$$\left( 16ac +\frac{a}{c^2b}+\frac{16c}{a^2d}+\frac{4}{ac}\right)\left( bd +\frac{b}{256d^2c}+\frac{d}{b^2a}+\frac{1}{64bd}\right) \ge \frac{81}{4}$$
When does the equality hold?
Given that \(a, b, c,\) and \(d\) are positive real numbers such that \(abcd = \frac{1}{4}\), we need to prove that:
\[
\left( 16ac +\frac{a}{c^2b}+\frac{16c}{a^2d}+\frac{4}{ac}\right)\left( bd +\frac{b}{256d^2c}+\frac{d}{b^2a}+\frac{1}{64bd}\right) \ge \frac{81}{4}
\]
and determine when equality holds.

1. **Substitution and Simplification:**
   Let \(ac = x\). Then, we can rewrite the given expression in terms of \(x\):
   \[
   \left( 16x + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{x} \right) \left( bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd} \right)
   \]

2. **Applying AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   16ac + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{ac} \geq 4 \sqrt[4]{16ac \cdot \frac{a}{c^2b} \cdot \frac{16c}{a^2d} \cdot \frac{4}{ac}}
   \]
   Simplifying the product inside the root:
   \[
   16ac \cdot \frac{a}{c^2b} \cdot \frac{16c}{a^2d} \cdot \frac{4}{ac} = 16 \cdot \frac{a^2}{c^2b} \cdot \frac{16c^2}{a^2d} \cdot \frac{4}{ac} = \frac{1024a^2c^2}{a^2c^2bd} = \frac{1024}{bd}
   \]
   Since \(abcd = \frac{1}{4}\), we have \(bd = 4ac\). Substituting \(bd = 4x\):
   \[
   \frac{1024}{bd} = \frac{1024}{4x} = \frac{256}{x}
   \]
   Therefore:
   \[
   16ac + \frac{a}{c^2b} + \frac{16c}{a^2d} + \frac{4}{ac} \geq 4 \sqrt[4]{\frac{256}{x}} = 4 \cdot \sqrt[4]{256} \cdot \sqrt[4]{\frac{1}{x}} = 4 \cdot 4 \cdot \sqrt[4]{\frac{1}{x}} = 16 \sqrt[4]{\frac{1}{x}}
   \]

3. **Simplifying the Second Term:**
   Similarly, for the second term:
   \[
   bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd} \geq 4 \sqrt[4]{bd \cdot \frac{b}{256d^2c} \cdot \frac{d}{b^2a} \cdot \frac{1}{64bd}}
   \]
   Simplifying the product inside the root:
   \[
   bd \cdot \frac{b}{256d^2c} \cdot \frac{d}{b^2a} \cdot \frac{1}{64bd} = \frac{b^2d^2}{256d^2c \cdot b^2a \cdot 64bd} = \frac{1}{256 \cdot 64 \cdot ac} = \frac{1}{16384ac}
   \]
   Since \(ac = x\):
   \[
   \frac{1}{16384ac} = \frac{1}{16384x}
   \]
   Therefore:
   \[
   bd + \frac{b}{256d^2c} + \frac{d}{b^2a} + \frac{1}{64bd} \geq 4 \sqrt[4]{\frac{1}{16384x}} = 4 \cdot \sqrt[4]{\frac{1}{16384x}} = 4 \cdot \frac{1}{8} \cdot \sqrt[4]{\frac{1}{x}} = \frac{1}{2} \sqrt[4]{\frac{1}{x}}
   \]

4. **Combining the Results:**
   Combining the two inequalities:
   \[
   \left( 16 \sqrt[4]{\frac{1}{x}} \right) \left( \frac{1}{2} \sqrt[4]{\frac{1}{x}} \right) = 8 \left( \sqrt[4]{\frac{1}{x}} \right)^2 = 8 \cdot \sqrt{\frac{1}{x}} = 8 \cdot \frac{1}{\sqrt{x}}
   \]
   Since \(x = 1\) when \(a = 2, b = 1, c = \frac{1}{2}, d = \frac{1}{4}\), we have:
   \[
   8 \cdot \frac{1}{\sqrt{1}} = 8
   \]
   Therefore, the inequality holds:
   \[
   \left( 16ac +\frac{a}{c^2b}+\frac{16c}{a^2d}+\frac{4}{ac}\right)\left( bd +\frac{b}{256d^2c}+\frac{d}{b^2a}+\frac{1}{64bd}\right) \ge \frac{81}{4}
   \]

The final answer is \( \boxed{ \frac{81}{4} } \)