First, strict inequalities make a problem. To illustrate it, let us consider
Minimize[{x, 1 < x && x < 2}, x]
Minimize::wksol: Warning: there is no minimum in the region in which the objective function is defined and the constraints are satisfied; returning a result on the boundary.
{1,{x->1}}
Second, the problem is too complicated for current CASes. To demostrate it, let us consider a simplified problem with weak inequalities instead of strict ones:
{(4 d k (1 - 2 r) r + k^2 r (-1 + 3 r) + d^2 (1 + r (-5 + 6 r)))/(2 (-1 + r)^2 r) +
q*(d^2 (2 - 3 r)+d k (-1 + 2 r) + k^2 (1+(-3 + r) r))/(-1 + r)^2, (2 r - 1)/r d <= k <= d}/.
{d -> 1/2, q -> 3/2};
Minimize[%, r]
which produces a piecewise expression in k
including
$$ \text{Root}\left[2048 \text{$\#$1}^3+\text{$\#$1}^2 \left(-7168 k^2+2048 k-3328\right)+\text{$\#$1} \left(8192 k^4-5120 k^3+9344 k^2-2304 k+1504\right)-3072 k^6+3072 k^5-6784 k^4+2880 k^3-2548 k^2+516 k-117\&,2\right]$$
Addition. The following
Minimize[ClearAll[d, k, q, r]; {(4 d k (1 - 2 r) r + k^2 r (-1 + 3 r) +
d^2 (1 + r (-5 + 6 r)))/(2 (-1 + r)^2 r) +
q*(d^2 (2 - 3 r) + 2 d k (-1 + 2 r) +
k^2 (1 + (-3 + r) r))/(-1 + r)^2, (2 r - 1)/r d <<= k <<= d &&
q r > 1/2 && r < 1} /. {d -> 1/2, q -> 3/2};Minimize[%, r]
works. The result is a piecewise expression which is too long to be cited here.
Edit. Missed inequalities r > 1/2 && r < 1
are added.