I would like to propose a generalization of another question which I posed here yesterday. The main reason is the heuristic that if an inequality holds for the finite case, then an integral analogue usually exists..
Let $(X,\Sigma,\mu)$ be a probability space and let $f:X\to \mathbf{R}$ be a measurable nonnegative function.
Then, is it true that $$ \begin{pmatrix} \int f^4 \mathrm{d}\mu & \int f^3\mathrm{d}\mu & \int f^2\mathrm{d}\mu \\ \int f^3\mathrm{d}\mu & \int f^2\mathrm{d}\mu & \int f\mathrm{d}\mu \\ \int f^2\mathrm{d}\mu & \int f\mathrm{d}\mu & 1 \\ \end{pmatrix} $$ has a nonnegative determinant?
[I think I can prove the claim in the case $(X,\Sigma,\mu)=([0,1],\mathscr{B}[0,1],\lambda \upharpoonright [0,1])$), i.e. with the Borel $\sigma$-field, and the Lebesgue measure; but it doesn't help much with this question, except suggesting the answer could be affirmative]
If $X_1$, $X_2$, $X_3$ are independent random variables with the distribution of $f$, i.e. $\mathbb E[X_i^j] = \int f^j\; d\mu$, we can write the determinant as $$ \eqalign{ \det &\pmatrix{ \mathbb E[X_1^4] & \mathbb E[X_2^3] &\mathbb E[X_3^2]\cr \mathbb E[X_1^3] & \mathbb E[X_2^2] & \mathbb E[X_3] \cr \mathbb E[X_1^2] &\mathbb E[X_2] & 1\cr} = \mathbb E \det \pmatrix{X_1^4 & X_2^3 & X_3^2\cr X_1^3 & X_2^2 & X_3\cr X_1^2 & X_2 & 1\cr} \cr &= \mathbb E \left[{X_{{1}}}^{4}{X_{{2}}}^{2}-{X_{{1}}}^{4}X_{{2}}X_{{3}}-{X_{{1}}}^{3}{X _{{2}}}^{3}+{X_{{1}}}^{3}X_{{2}}{X_{{3}}}^{2}+{X_{{1}}}^{2}{X_{{2}}}^{ 3}X_{{3}}-{X_{{1}}}^{2}{X_{{2}}}^{2}{X_{{3}}}^{2} \right]} $$ Averaging over permutations of $X_1, X_2, X_3$, this becomes $$\dfrac{1}{6} \mathbb E[(X_1 - X_2)^2 (X_1 - X_3)^2 (X_2 - X_3)^2] $$ which is of course nonnegative.