Without loss of generality, we can assume that $ a \ge c. $ Since $ a, b, d <1 $, by the Cauchy inequality we have $$(ab-cd)(ad-bc)\le \left(\frac{(ab-cd)+(ad-bc)}{2}\right)^ 2=\frac{(b+d)^2(a-c)^2}{4} < (1-c)^2, \quad (1)$$and $$ ac + bd <c + 1. \quad (2) $$Multiplying the inequalities $(1)$ and $(2)$, we get: $$ (ab-cd) (ac + bd) (ad-bc) <(1-c) ^ 2 (1 + c) = ( 1-c) (1-c ^ 2) <1-c \le 1- \min (a, b, c, d). $$