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

    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) + d k (-1 + 2 r) + k^2 (1 + (-3 + r) r))/(-1 + r)^2, (2 r - 1)/r d <= k <= d && 
     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.