Consider hermitian positive semi-definite matrices $A\_1$ and $A\_2$. Consider also positive definite matrices $B\_1$ and $B\_2$. I want to maximize the minimum of the two Generalized Rayleigh Ritz ratios $\frac{x^{H}A\_1x}{x^{H}B\_1x}$ and $\frac{x^{H}A\_2x}{x^{H}B\_2x}$. To state it formally, the problem is 


\begin{align}
\max\_{x} \min\left(\frac{x^{H}A\_1x}{x^{H}B\_1x},\frac{x^{H}A\_2x}{x^{H}B\_2x}\right)
\end{align}


I would be very much happy if we can find a closed form solution for it. I would also appreciate if some one can suggest a iterative algorithm to solve it. I am not interested in converting it to a convex optimization problem even if it is possible. (But, if convertible to a convex is fruitful, then I would like to see it). 


EDIT------ 


My attempt on the problem 


Introduce a extra variable $t$ and rewrite the problem as 


\begin{align}
\max\_{x,t}~t , 
\end{align}
\begin{align}sucht~that~
\frac{x^{H}A\_1x}{x^{H}B\_1x} \geq t,
\frac{x^{H}A\_2x}{x^{H}B\_2x} \geq t
\end{align}
Since $B\_1$ and $B\_2$ are positive definite, one can re-write this equations as 


\begin{align}
\max\_{x,t}~t , 
\end{align}
\begin{align}sucht~that~
x^{H}(A\_1-t B\_1)x \geq 0
\end{align}
\begin{align}
x^{H}(A\_2-t B\_2)x \geq 0 
\end{align}


May some one here can come up with a algorithm for this problem. 



Simultaneous maximization of two Generalized Rayleigh Ritz Ratios
Here is a crude idea that might work (haven't thought too carefully about it).


Let $a=\lambda\_{\min}(B\_1^{-1}A\_1)$ and $b=\lambda\_{\min}(B\_2^{-1}A\_2)$. Then, for there to be a feasible solution to the 2nd formulation, the variable $t$ must lie in the interval $[0,t\_{\max}]$, where $t\_{\max} := \min(a,b)$. 


This suggests that maybe we can do a *binary-search* for the optimal value of $t$.


1. Start with $t = t\_{\max}/2$
2. With this value of $t$, either there is a feasible $x$ that satisfies both inequalities, or there isn't. If there isn't, we shrink $t$, else we expand it; that is, if there is no feasible $x$, we search in $[0,t]$, otherwise we search in $[t, t\_{\max}]$ and repeat this step.


(Notice that we obtain an optimal $x$ as a by-product).


