the minmum value of 96

By adding the two sums-of-squares equations above, one obtains:

$a^2 + b^2 +c^2 + d^2 = 2$ (i)

and after adding $2ac + 2bd$ to either side of (i) we get:

$(a^2 + 2ac + c^2) + (b^2 + 2bd + d^2) = 2 + 2ac + 2bd \Rightarrow \frac{(a+c)^2 + (b+d)^2 - 2}{2} = \frac{1}{2}[(a+c)^2 + (b+d)^2] - 1 = ac + bd$ (ii).

and after subtracting 2 from either side of (ii) we finally obtain:

$\frac{1}{2}[(a+c)^2 + (b+d)^2] - 3 = ac + bd - 2$ (iii).

Since $(a+c)^2 + (b+d)^2 \ge 0$ , the minimum value computes to $ac+bd-2 = \boxed{-3}.$

Let $a=\cos α,c=\cos β$ . Then $b=\sin α,d=\sin β$

So $ac+bd-2=\cos (α-β)-2\geq -3$ (since $\cos (α-β)\geq -1$ )

Hence the required minimum is $\boxed {-3}$ when $ac+bd=-1$ .