This question is, essentially, a restatement of the previous question. Now that the previous question is answered, the question on this page can be answered as well.
Indeed, let \begin{equation} W:=\sum_i b_i u_i, \end{equation} where $u_i$ is the $i$th column of the matrix $G^{1/2}$. Then \begin{equation} W^\top W=b^\top Gb>0, \end{equation} because the matrix $G$ is positive definite and $(Gb)_i>0$ for all $i$, so that $b\ne0$. So, we can introduce the unit vector \begin{equation} w:=\frac W{(W^\top W)^{1/2}}=\frac W{(b^\top Gb)^{1/2}}, \end{equation} and then we will have $u_j^\top W=(Gb)_j>0$ and hence $u_j^\top w>0$ for all $j$.
Therefore, by the mentioned answer, \begin{equation} S:=\sum_i |u_i^\top a| u_i^\top w<1. \end{equation} On the other hand, \begin{equation} S=\frac1{(b^\top Gb)^{1/2}}\sum_j |u_j^\top a| u_j^\top W =\frac1{(b^\top Gb)^{1/2}}\sum_{j,i} b_j b_i G_{ij}=(b^\top Gb)^{1/2}. \end{equation} Thus, $b^\top Gb<1$. $\quad\Box$