Let $ A $ be an $n\times n$ symmetric matrix. Then its eigenvalues are all real. Is it possible to express the fact that all eigenvalues $ c_1, \dots, c_n $ satisfy the same inequality, say, $ a < c_k < b $, in terms of the entries of $ A $?
Because the general equation of $ n $-th degree is not solvable for $ n \geq 5 $, it is impossible to express the characteristic roots of a general matrix in terms of its entries, but I am essentially asking if this is possible if one already knows that the characteristic polynomial has only real roots. Note that the inequalities above are invariant under permutations of $ \{ 1,\dots,n\}$, a fact which may be relevant.
This is the same as saying that $aI< A< bI$ where $A<B$ means that $B-A$ is positive definite. A symmetric matrix is positive definite iff each matrix formed by intersection of the first $k$ rows and first $k$ columns has positive determinant for all $k$ from $1$ to $n$. So we can express your condition in terms of a finite set of strict inequalities.