Check this out: http://en.wikipedia.org/wiki/Fisher_information#Matrix_form
From the definition, we have
$$
I_{ij} = \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] \, ,
$$
for $i,j=1,\dots,k$, in which $\partial_i=\partial /\partial \theta_i$. Your expression for $I_{ij}$ follows from this one under regularity conditions.
For a nonnull vector $u = (u_1,\dots,u_k)^\top\in\mathbb{R}^n$, it follows from the linearity of the expectation that
$$
\sum_{i,j=1}^k u_i I_{ij} u_j = \sum_{i,j=1}^k \left( u_i \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] u_j \right) \\
= \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\sum_{j=1}^k u_j \partial_j \log f_{X\mid\Theta} (X\mid\theta)\right)\right] \\
= \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right)^2 \right] \geq 0 \, .
$$
If this component wise notation is too ugly, note that the Fisher Information matrix $H=(I_{ij})$ can be written as $H = \mathrm{E}_\theta\left[S S^\top\right]$, in which the scores vector $S$ is defined as
$$
S = \left( \partial_1 \log f_{X\mid\Theta}(X\mid\theta), \dots, \partial_k \log f_{X\mid\Theta}(X\mid\theta) \right)^\top \, .
$$
Hence, we have the one-liner
$$
u^\top H u = u^\top \mathrm{E}_\theta[S S^\top] u = \mathrm{E}_\theta[u^\top S S^\top u] = \mathrm{E}_\theta\left[|| S^\top u ||^2\right] \geq 0.
$$