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If $A=(a_{ij})$ is a real positive definite symmetric matrix of order $n$. How to show $(n-1)\prod_{i=1}^na_{ii}+\det A\ge \sum_{i=1}^na_{ii}\det A(i)$? $A(i)$ means the submatrix of $A$ by deleting the $i$th row and $i$th column.

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if $n=3$. consider the two matrix below:

$$A=\begin{pmatrix}1 & \frac{1}{4} & -\frac{1}{4}\cr \frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$

$$B=\begin{pmatrix}1 & -\frac{1}{4} & -\frac{1}{4}\cr -\frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$

they are both positive definite matrix. but for matrix $A$, we have:$$2\prod_{i=1}^3a_{ii}+\det A> \sum_{i=1}^3a_{ii}\det A(i)$$ hold.

but for matrix $B$, the inequality don't hold. and matrix $B$ satisfy:$$2\prod_{i=1}^3b_{ii}+\det B< \sum_{i=1}^3b_{ii}\det B(i)$$

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I guess the proposer missed a condition $n\ge 4$. –  Fischer Jun 1 '13 at 13:17

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