Given positive reals $a,b$, and $c$, is it true that the matrix $$ \begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix} $$ has nonnegative determinant?
I conjecture the answer is affirmative; on the other hand, at the moment I didn't try with numerical evidences..
Note that $$ \begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix}=\begin{pmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1\\ \end{pmatrix}\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\\ \end{pmatrix} $$ The determinant thus is the square of the determinant of $$\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\\ \end{pmatrix} $$ which is $(a-b)^2(b-c)^2(c-a)^2$ which is non negative for all real $a,b,c$.