Proving the determinant of a matrix is positive How can I show that $\det  A \leq 0 $,
if $A = \begin{pmatrix}
a & b & c\\ 
b & d & e\\ 
c & e & f
\end{pmatrix}$ 
and $a>0, ad-b^2 = 0$

We have $$|A|=adf-b^2f-ae^2-dc^2+2bce=-(ae^2+dc^2-2bce)=-(ae^2+dc^2\pm 2\sqrt{ad}ce)=-(e\sqrt{a}\pm c\sqrt{d})^2\le 0$$
Note that we need $a>0$ for the step  $\sqrt{ad}=\sqrt{a}\sqrt{d}$.
