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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.

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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.

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    $\begingroup$ More briefly: $a^T G a \leq a^T a \leq (a^T G a)^{1/2} (a^T G^{-1} a)^{1/2}$, hence $a^T G a \leq a^T G^{-1} a = 1$. $\endgroup$
    – Terry Tao
    Commented Jun 5, 2023 at 3:07
  • $\begingroup$ This simple? :-) Thank you for your answer. $\endgroup$
    – Iosif Pinelis
    Commented Jun 5, 2023 at 3:14
  • $\begingroup$ @TerryTao : This is is even simpler! Thank you for your comment. $\endgroup$
    – Iosif Pinelis
    Commented Jun 5, 2023 at 3:16

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