Very good one.

parmenides51 wrote: 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? Let $ac=x$. Thus, by AM-GM we obtain: $$\left( 16ac +\frac{a}{c^2b}+\frac{16c}{a^2d}+\frac{4}{ac}\right)\left( bd +\frac{b}{256d^2c}+\frac{d}{b^2a}+\frac{1}{64bd}\right)\geq$$ $$\geq\left( 16ac +\frac{8}{\sqrt{abcd}}+\frac{4}{ac}\right)\left( bd +\frac{1}{8\sqrt{abcd}}+\frac{1}{64bd}\right)=$$$$=\left(16x+16+\frac{4}{x}\right)\left(\frac{1}{4x}+\frac{1}{4}+\frac{x}{16}\right)=\frac{(2x+1)^2(x+2)^2}{4x^2}=$$$$=\frac{1}{4}\left(2\left(x+\frac{1}{x}\right)+5\right)^2\geq\frac{1}{4}\left(2\cdot2+5\right)^2=\frac{81}{4}.$$The equality occurs for $$(a,b,c,d)=\left(2,1,\frac{1}{2},\frac{1}{4}\right).$$

In this problem, its very difficult to guess the equality cases. Usually, when I do an inequality, I try to make sure that all the equality cases hold at every step of the way to make sure that the steps I am making are sharp, but I can only do that when I know all(or at least some) of the equality cases to guide me. In most problems, it is easy to find at least one equality cases, such as making all the variables equal or setting some of them to extremes while others to 0, but in this equality, the equality case seems rather random. Is there any "natural" way to see the equality case in this problem?