Below, $a_i$ is a column vector with matching dimension as the square matrix $G$ and $b_{ij}$ is a scalar. I apply CS once over $(i,j)$ to obtain the first inequality but I'd like to apply CS once more to the first term on the right hand side, to basically decompose the bound into the Euclidean norm of the matrix $G$ and the norm of the vector $a$.
$$\sum_i^N \sum_j^N a_{i}'\,G \,a_{j} \cdot b_{ij} \leq \left(\sum_i^N\sum_j^N \left[ a_{i}'G a_{j}\right]^2\right)^{\frac{1}{2}} \cdot \left(\sum_i^N\sum_j^N b_{ij}^2\right)^{\frac{1}{2}}$$
Any suggestions?
Here is how I would do. Notice that if $x$ and $y$ are two vectors we have by Cauchy-Schwarz
$$ \begin{aligned} (x^T G y)^2 &= \left(\sum_{k,l} G_{kl} x_k y_l\right)^2 \leq \underbrace{\left(\sum_{k,l} G_{kl}^2 \right)}_{\|G\|^2} \left(\sum_{k,l} x_k^2 y_l^2 \right) \\ &= \|G\|^2 \left(\sum_k x_k^2 \right) \left(\sum_l y_l^2 \right) = \|G\|^2 \|x\|^2 \|y\|^2 \end{aligned} $$
so applying this with $x = a_i$, $y = a_j$ in your case you get
$$ \sum_{i,j} (a_i^T G a_j)^2 \leq \sum_{i,j} \|G\|^2 \|a_i\|^2 \|a_j\|^2 = \|G\|^2 \left(\sum_i \|a_i\|^2\right)^2 = \|G\|^2 \|a\|^4 $$
so
$$ \left(\sum_{i,j} (a_i^T G a_j)^2 \right)^{1/2} \leq \|G\| \|a\|^2 $$
if you consider $a$ as a matrix or a long vector for the last equality. Hope this helps!