Positive determinant of a symmetric matrix Let $a,b,c,d,e$ positive reals for which $b<\sqrt{ac}$, $c<\sqrt{bd}$, and $d<\sqrt{ec}$. Then, consider the following (Hankel) matrix
$$
M=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\\ \end{pmatrix}
$$ 
Is it possible to prove that the determinant of $M$ is positive by means of "symmetric" arguments?
 A: To facilitate the computation of the determinant, remember it stays the same if you add one row/column to another row/column times a constant. So given the shape of the matrix you can do this process to "clean it out".
1) substract to column 1 the column 2, afterwards substract to column 2 the column 3.
2) subtract to row 1 the row 2, then substract to row 2 the row 3.
3) substract to row 1 the row 2, and to column 1 the column 2.
You will end up with matrix:
$$ \left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right) $$
its determinant is 
$$ e(ac-b^2) - d(da-bc) + c(db-c^2) $$
A: The claim is not true. The conditions on the coefficients ensure that certain submatrices of the matrix 
$$
\left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right)
$$
are positive definite. These conditions are not sufficient to ensure that the whole matrix is positive definite.
Here is a counter-example: choose $a=e=5$, $b=d=2$, $c=1$. Then the conditions on the coefficients are satisfied, however the matrix
$$
\left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right)
=
\left( \begin{array}{ccc} 5 & 2 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & 5 \end{array} \right)
$$
has determinant $-8$.
