= Solution
Use the <optimization Lagrangian>
$$
L=x^TAx+y^TBy-\lambda(\mathbf1^Tx+\mathbf1^Ty-1).
$$
Symmetry gives the <gradient> equations $2Ax=\lambda\mathbf1$ and $2By=\lambda\mathbf1$. The <positive-definite matrices> are invertible, so define
$$
D=\mathbf1^TA^{-1}\mathbf1+\mathbf1^TB^{-1}\mathbf1>0.
$$
The affine constraint gives $\lambda=2/D$, hence
$$
\boxed{x_* =\frac{A^{-1}\mathbf1}{D},\qquad
y_* =\frac{B^{-1}\mathbf1}{D},\qquad f_* =\frac1D.}
$$
For a global certificate, write any other feasible point as $(x_*+u,y_*+v)$. Feasibility implies $\mathbf1^Tu+\mathbf1^Tv=0$. Expansion cancels the cross terms and leaves
$$
f(x_*+u,y_*+v)-f(x_*,y_*)=u^TAu+v^TBv.
$$
By positive definiteness this is positive unless $u=v=0$. Thus the solution is the \b[unique <global minimum>], not merely a <stationary point>. This is a two-block instance of <positive quadratic minimization under one affine constraint>.
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