This question is, essentially, a restatement of the [previous question][1]. Now that the previous question is [answered][2], the [question on this page][3] 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][2], 
\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, $b^\top Gb<1$. $\quad\Box$ 


[1]: https://mathoverflow.net/q/446652/36721 
[2]: https://mathoverflow.net/a/448631/36721 
[3]: https://mathoverflow.net/q/448364/36721