 Solution

Use the optimization LagrangianSymmetry gives the gradient equations $2Ax=\lambda 1$ and $2By=\lambda 1$. The positive-definite matrices are invertible, so defineThe affine constraint gives $\lambda =2/D$, henceFor a global certificate, write any other feasible point as $(x_{\ast }+u,y_{\ast }+v)$. Feasibility implies $1^{T}u+1^{T}v=0$. Expansion cancels the cross terms and leavesBy positive definiteness this is positive unless $u=v=0$. Thus the solution is the unique global minimum, not merely a stationary point. This is a two-block instance of positive quadratic minimization under one affine constraint.