Let $\vec{a}$ and matrix $R$ (square positive definite triangular) be given. I need to restructure this expression
$$ P_R(\vec{a})=\sum_\vec{b} \frac{1}{(a\circ b)^HR^HR(a\circ b)} $$
where the sum is over several different $\vec{b}$. I need to do this calculation for several different $\vec{a}$, so I want to find some expression in the form of
$$ P_R(\vec{a}) = f(\vec{a}) $$
where $f$ contains some pre-calculated values from the sum. I.e. I want to find an sub-expression which I can use for every $\vec{a}$.
I'm having problems with the reciprocal. If it were not there, I could do the following: $$ \begin{align} P_R(\vec{a})&=\sum_\vec{b} (a\circ b)^HR^HR(a\circ b)\\ &= tr(B^H\text{diag}(\vec{a}) R^HR \text{diag}(\vec{a})^HB) \end{align} $$
where $B$ contains as the columns all values of $\vec{b}$. Now, I can introduce unitary Fourier matrices into beginning and end of trace, to get as the $j$th column of $$ (FB^H\text{diag}(\vec{a}))_{:,j}=a_j F\vec{b}_j $$ where $a_j$ is the $j$th element of $\vec{a}$ and $\vec{b}_j$ is the $j$ the column of $B$. Hence, $$ FB^H\text{diag}(\vec{a}) = \text{diag}(\vec{a})FB^H $$
And the initial expression can be formulated to
$$ P_R(\vec{a})=tr(\text{diag}(\vec{a})\text{diag}(\vec{a}) FB^HR^HRBF^H) $$
which contains an expression which is independent of $\vec{a}$ and hence I can pre-calculate it. However, unfortunately there is this reciprocal which I cannot overcome. I've tried going to the log (or exponential) domain etc. but did not find a good solution.