Does Generalized Eigenvector solve this problem? Let $\mathbf{A}$ and $\mathbf{B}$ be two $N\times N$ hermitian matrix, then I need a non-zero vector $\mathbf{x}$ such that 
\begin{align}
\mathbf{x}^H\mathbf{A}\mathbf{x}\geq 0, \mathbf{x}^H\mathbf{B}\mathbf{x}\geq 0
\end{align}
Does the generalized eigen vectors  of $\mathbf{Ax=\lambda Bx}$ corresponding to positive eigenvalues satisfy them? if not, why? Is there any special situation under which it will be so? say what if $\mathbf{A}$ is positive definite?
 A: Let $A = U \Lambda_A U^*$, $\Lambda_A = {\rm diag}(\lambda^A_1,\dots,\lambda^A_n)$, and $B = V \Lambda_B V^*$, $\Lambda_B = {\rm diag}(\lambda^B_1,\dots,\lambda^B_n)$, be eigenvalue (Schur) decompositions of $A$ and $B$, respectively. Without the loss of generality, we can assume that
$$\lambda^X_i \ge \lambda^X_j, \quad \text{for $i < j$ and $X \in \{A,B\}$},$$
i.e., the diagonal elements of both $\Lambda$ are in nonincreasing order.
Let us denote the index of the last nonnegative diagonal element in $\Lambda_A$ by $k_A$, i.e., $\lambda^A_{k_A} \ge 0$ and $\lambda^A_{k_A} < 0$ (borderline cases, when all $\lambda^A_i$ are all nonnegative or all negative, being obvious). Similarly, define $k_B$ to be the index of the last nonnegative diagonal element in $\Lambda_B$.
Obviously,
$$u_i^* A u_i = u_i^* U \Lambda_A U^* u_i = e_i^* \Lambda_A e_i = \lambda^A_i, \quad v_i^* B v_i = v_i^* V \Lambda_B V^* v_i = e_i^* \Lambda_B e_i = \lambda^B_i,$$
where $e_i$ are vectors of the canonical basis (i.e., columns of the identity matrix). We conclude that
$$u_i^* A u_i \ge 0, \quad \text{for all $i \le k_A$}, \quad \text{and} \quad v_i^* B v_i \ge 0, \quad \text{for all $i \le k_B$}.$$
This means that your solutions $x$ (not all; see the comments) are elements of the intersection of the spaces induced by these (obviously nonlinear) sets, i.e.,
$$x \in {\rm span}\{u_1,\dots,u_{k_A}\} \cap {\rm span}\{v_1,\dots,v_{k_B}\}.$$
Finding the intersection of two spaces given by their orthonormal bases should not be a problem.
Extra case
This is related to the comments below, where you stated that the above does not fit your needs.
Note that
$$u_i^* A u_j = u_i^* U \Lambda_A U^* u_j = e_i^* \Lambda_A e_j = 0, \quad \text{for $i \ne j$}.$$
Let $i, j$ be such that $\lambda^A_i = u_i^* A u_i > 0$ and $\lambda^A_j = u_j^* A u_j < 0$. Then
$$(\alpha u_i + u_j)^* A (\alpha u_i + u_j) = |\alpha|^2 u_i^* A u_i + 2{\rm Re}(\overline{\alpha} u_i^* A u_j) + u_j^* A u_j = |\alpha|^2 u_i^* A u_i + u_j^* A u_j = 0$$
for some real $\alpha$, which we denote $\alpha_{ij}$. Furthermore, denote
$$x_{ij}(\alpha) := \frac{1}{\sqrt{|\alpha u_i + u_j|}} (\alpha u_i + u_j).$$
Now, we define
$$S_A := \{ t x_{ij}(\alpha)\!:\ \lambda^A_i > 0,\ \lambda^A_j < 0,\ \alpha \ge \alpha_{ij},\ t \ge 0 \}.$$
Note that $S_A$ is a cone given by the unitary generators $x_{ij}(\alpha_{ij})$ and, for $x \in S_A$, we have $x^* S_A x \ge 0$.
Define $S_B$ in a similar manner.
Now, $S_A$ and $S_B$ are cones given by unitary generators, satisfying  your question's first and second inequality, respectively. So, the vectors in the intersection $S_A \cap S_B$ are the vectors you're looking for. If needed, this could be combined with the above vector space to produce the cone of all solutions (this happens if $S_A \cap S_B$ spans a proper subspace of $\mathbb{C}^N$), but you said you're not interested in that.
There remains the problem of finding an intersections of two cones. I am not an expert, but given the set of unitary generators, I expect there is a known algorithm for this.
