Consider two real $n \times n$ symmetric matrices $A$ and $B$, where $n$ is a given positive integer. Write the elements in $A$ and $B$ as $\{a_{ij}\}$ and $\{b_{ij}\}$, where $1 \leq i \leq n$ and $1 \leq j \leq n$. Assume that $a_{ii}=b_{ii}=1$ for $1 \leq i \leq n$ and $0 \leq a_{ij} \leq b_{ij} \leq 1$ for all $1 \leq i<j \leq n$. We want to show the following:
If $B$ is positive-definite, then $A$ is also positive-definite.
Actually, from Sylvester's criterion, we can reduce the problem to prove the following (This is equivalent to our initial question):
If the determinant of $B$ is a positive real number, then the determinant of $A$ is also a positive real number.
Thus, for $n=2$ case, the proof is just a direct calculation. However, for $n \geq 3$, the calculation becomes difficult and it is hard to prove the result in general.
Any ideas and comments on the problem are very welcome. Thanks a lot!
Both statements are false. Let $$ B=\pmatrix{1&b&b\\ b&1&b\\ b&b&1}, \ A=\pmatrix{1&b&b\\ b&1&a\\ b&a&1} $$ where $b<1$ is close to $1$ and $a>0$ is close to $0$. Then $B=(1-b)I+bee^T$ is positive definite, but $\det(A)=1+2ab^2-a^2-2b^2\to1-2b^2$ is negative when $a\to0^+$.