Linear elliptic partial diff. operator Let $L$ be the above operator 
\begin{equation}
L u  = \sum_{i,j=1}^{n} a_{ij}(x) u_{x_ix_j} + \sum_{i=1}^{n} \beta_i u_{x_i} + c(x) u(x)
\end{equation}
Assume furthermore that L is elliptic i.e. 
\begin{equation}
0 < \lambda {|\vec{\xi }|}^2 \leq \sum_{i,j=1}^{n} a_{ij}\xi_i\xi_j \text{ for every } \vec{\xi} \in \mathbf{R}^n\setminus\{ \vec{0}\} \text{and for every x in the domain }   
\end{equation}
The last relation yields that the matrix $[A]_{ij} = a_{ij}$ is positive definite.
So here's the thing if at the point $x_0$ we have a maximum then the matrix of the second derivative of $u$ at point $x_0$ is negative semi-definite, so (hessian) $[H]_{ij}=u_{x_ix_j}(x_0)$ and $x^t H x \leq 0 $.
How can we conclud that
\begin{equation}
\sum_{i,j=1}^{n} a_{ij}(x_0) u_{x_ix_j} (x_0) \leq 0
\end{equation}
I know that the product of matrix A and the H is negative semi definite (it is  easy to show the last one). 

If you already know how to show that $AH$ is negative semi-definite, then just note that 
$$\sum_{i,j=1}^n a_{ij}u_{x_ix_j} = \text{Trace}(AH)\leq 0.$$
