Positive values for a set of quadratic forms of Hermitian Matrices. (To find a set of vectors in which a hermitian matrix is positive definite) Assume all matrices I discuss about are $N \times N$ and the vectors conform with dimensions. Consider the following set of Quadratic inequalities where all the matrices $A_i$ are hermitian. 
\begin{align}
x^{H}A_{1}x &\geq 0 \\
x^{H}A_{2}x &\geq 0 
\end{align}
I need to find at least one non-zero vector $x$ such that these inequalities are satisfied at the same time. In fact, the method need not be constructive. I just need to prove, there exist at least one vector which satisfy this equation. Also each $A_i$ is guaranteed to have at least one negative eigen value and at least one positive eigen value, i.e. they are indefinite matrices. Also $A_1$ and $A_2$ are strictly complex. ( You can find examples for real $A_1$ and $A_2$ for which this won't hold, thanks to user @Marc).  
EDIT---- MY ATTEMPT
Consider the optimization problem
\begin{align}
\max_{x,t}~~~t \\
subject~to~ x^{H}A_{1}x & \geq t \\
x^{H}A_{2}x & \geq t 
\end{align}
which is equivalent to
\begin{align}
\min_{x,t}~~~-t \\
subject~to~ x^{H}A_{1}x & \geq t \\
x^{H}A_{2}x & \geq t 
\end{align}
If it is feasible, this problem will tell me (in some sense the best) an $x$ for the problem possed initially. Consider the lagrangian
\begin{align}
L(x,t,u)= -t + \lambda_{1}(t-x^{H}A_{1}x)+\lambda_{2}(t-x^{H}A_{2}x)
\end{align}
where $u$ is vector containing the lagrangian multipliers $\lambda_1$ and $\lambda_2$. The lagrangian dual function is given by 
\begin{align}
h(u)= \max_{x,t}~L(x,t,u)
\end{align}
Now its known that the lagrangian dual function $h(u)$ for all $u\geq 0$ is a lower bound for the original objective function. Hence if we can find some $u$ (i.e. $\lambda_1$ and $\lambda_2$ such that $h(u)=0$, then the original objective function would be greater than 0 and hence that particular $x$ will be the one that satisfies all the inequalities. Now the differentiating the lagrangian dual function $h(u)$ w.r.to $x$ and $u$, one obtains the following conditions.
\begin{align}
\lambda_1+\lambda_2 &=1 \\
(\lambda_1A_1+\lambda_2A_2)x &= 0
\end{align}
Hence, I get the condition that for my problem to be solvable, there should exist a convex combination of $A_1$ and $A_2$ such that their null space is non-empty or equivalently they have a zero eigen value. This problem looks like a very fundamental question though not so obvious. Any way, I am going to pose this as a new problem. 

$$
  A_1=\begin{pmatrix}1&0\\0&-2\end{pmatrix} \quad\text{and}\quad
  A_2=\begin{pmatrix}-2&0\\0&1\end{pmatrix}.
$$
