Suppose that $G=(G\_{ij})$ is an $n\times n$ positive-definite symmetric matrix with the diagonal entries all equal $1$ and all off-diagonal entries $\le0$. Let $a$ be a column $n\times1$ matrix with nonnegative entries such that $a^\top G^{-1}a=1$. Does it then necessarily follow that $a^\top Ga\le1$?


Certain numerical experiments suggest that this is true.



An inequality for certain positive-definite matrices
The answer seems to be **yes**.


Let $G$ be the Gram matrix of a base $(e\_1,\dots,e\_n)$ in some Euclidean space. Then $G^{-1}$ is the Gram matrix of the dual base $(f\_1,\dots,f\_n)$, i.e., the one satisfying $\langle e\_i,f\_j\rangle=\delta\_{ij}$.


Denote $u=\sum\_i a\_ie\_i$ and $v=\sum\_j a\_jf\_j$; then $\|u\|^2=a^TGa$ and $\|v\|^2=a^TG^{-1}a$. . The properties of $G$ immediately yield that
$$
 \langle u,u\rangle=a^TGa=\sum\_ia\_i^2-(\text{nonnegative})\leq\sum\_ia\_i^2=\langle u,v\rangle.
$$
So $\|u\|\|v\|\geq \langle u,v\rangle\geq \|u\|^2,$ which yields $|v\|\geq \|u\|$, as desired.


