Let $A$ and $B$ be two given hermitian positive semi-definite matrices, then what is the solution for 


$$\max\_{x\neq 0}\frac{x^HAx}{x^HBx+1}.$$


I am looking for closed form solutions. 
If the denominator didn't have that $1$, this is standard generalized rayleigh quotient and would be unbounded.


I know how to solve it numerically. The trick is to re-write it as 


$$\max\_{x,t}~t\\ \text{s.t.}~~x^H(A-tB)x \ge t$$


Then find the largest $t$ such that there exists a $x$ which satisfies $x^H(A-tB)x\gt0$. A Bi-section search on $t$ will do the job.


I have already asked this question before in math stack exchange. Since I didn't get any answers, I thought I will post it up here. 



An Interesting variant of Rayleigh Quotient
Replacing any nonzero $x$ by $tx$ with real $t$,
$ \dfrac{(tx)^H A (tx)}{(tx)^H B (tx) + 1}$ increases to $\dfrac{x^H A x}{x^H B x}$ as $t \to \infty $. Thus the supremum is your generalized Rayleigh quotient. If $B$ is positive definite, the supremum (not maximum, as it is not attained) is the largest eigenvalue of $B^{-1/2} A B^{-1/2}$. If $\text{ker}(B)$ is not contained in $\text{ker}(A)$, it is
unbounded.


