Sylvester's criterion says that an $n\times n$ Hermitian matrix $A$ is positive definite if and only if all its leading principal minors of are positive. If one knows that fact that every Hermitian matrix has an orthogonal eigenbasis, one can prove Sylvester's criterion easily.
The forward implication is obvious. If $A$ is positive definite, so are all its leading principal submatrices. Their spectra are hence positive. Thus the leading principal minors are positive, because each of them is a product of the eigenvalues of the submatrix.
To prove the backward implication, we use mathematical induction on $n$. The base case $n=1$ is trivial. Suppose $n\ge2$ and all leading principal minors of $A$ are positive. In particular, $\det(A)>0$. If $A$ is not positive definite, it must possess at least two negative eigenvalues. As $A$ is Hermitian, there exist two mutually orthogonal eigenvectors $x$ and $y$ corresponding to two of these negative eigenvalues. Let $u=\alpha x+\beta y\ne0$ be a linear combination of $x$ and $y$ such that the last entry of $u$ is zero. Then
$$
u^\ast Au=|\alpha|^2x^\ast Ax+|\beta|^2y^\ast Ay<0.
$$
Hence the leading $(n-1)\times(n-1)$ principal submatrix of $A$ is not positive definite. By induction assumption, this is impossible. Hence $A$ must be positive definite.
If you want to find all possible values of off-diagonal elements of a matrix A such that A is positive definite and all the off-diagonal elements are negative:
First assume that the diagonal elements are given, $a_{11}, a_{22},a_{33},a_{44}$, which must be positive. I will presume that by positive definite, you mean symmetric positive definite. Let the 6 upper triangular elements be variables, a_{12},a_{13},a_{14},a_{23},a_{24},a_{34}. The lower triangle is determined by symmetry.
Using MATLAB notation, you need all principal minors to be positive:
det(A(1:2,1:2)) > 0 , det(A(1:3,1:3)) > 0, det(A) > 0 (note that det(A(1:1,1:1)) = $a_{11}$ which is > 0 by assumption). You can expand these out symbolically, which results in a set of 3 strict inequalities in 6 variables. Additionally. there are negativity constraints on all 6 variables. The set of all solutions to these 9 inequalities is the solution to your problem. However, from a practical perspective, this is not a good way to compute things, and really just serves as a theoretical exercise.
As for the question in your comment "can you please suggest a way to show that all such matrices, that is those with off diagonal elements negative but still positive definite form a convex set?": You should already know that the set of all positive definite matrices is convex (a convex cone). Any convex combination of negative numbers will be negative, therefore, a convex combination of matrices having negative off-diagonal elements will have negative off-diagonal elements. Hence the set of all matrices having negative off-diagonal elements is convex. Hence, the set of all positive definite matrices having negative off-diagonal elements is the intersection of two convex sets, hence is convex.
Best Answer
If $D$ is a diagonal matrix with $(b_1,\ldots,b_n)$ in the diagonal and $b_i>0$ then $tr(AD)=a_{11}b_1+\ldots a_{nn}b_n$. Since $a_{ii}>0$ and $b_i>0$ then $a_{11}b_1+\ldots +a_{nn}b_n\leq(a_{11}+\ldots +a_{nn})(b_1+\ldots+b_n)=tr(A)tr(D)$.
If $B$ is symmetric then $B=RDR^t$, where $R$ is orthogonal and $D$ is diagonal, and $tr(AB)=tr(ARDR^t)=tr(R^tARD)\leq tr(R^tAR)tr(D)=tr(A)tr(B)$.