Is there a way to prove that $$ S = \left(\,\left\langle\, x^{4}\,\right\rangle - \left\langle\, x^{2}\,\right\rangle^{2}\,\right) \left(\,\left\langle\, x^{2}\,\right\rangle - \left\langle\, x\,\right\rangle^{2}\,\right) - \left(\,\left\langle\, x^{3}\,\right\rangle - \left\langle\, x^{2}\,\right\rangle \left\langle\, x\,\right\rangle\,\right)^{2}\ >\ 0 $$ for a real random variable defined over $-1\ \leq\ x\ \leq\ 1$ with any distribution function $\,\mathrm{f}\left(\, x\, \right)$ whose support is at least three distinct points ?. This inequality seems to hold for all distribution functions I tried. I am even not sure if the $-1\ \leq\ x\ \leq\ 1$ condition is necessary, I encountered this expression in physics research where $x\equiv \cos\left(\,\theta\,\right)$.
Here the expectation value is defined in the usual way
$$\left\langle\, \mathrm{y}\,\right\rangle \equiv \frac{\displaystyle{\int_{-1}^{1}\mathrm{y}\left(\, x\,\right)\,\mathrm{f}\left(\,x\,\right)\, \mathrm{d}x}} {\displaystyle{\int_{-1}^{1}\mathrm{f}\left(\, x\,\right)\,\mathrm{d}x}} $$
for any $\,\mathrm{y}\left(\, x\,\right)$ function where $\,\mathrm{f}\left(\, x\,\right) \geq 0$ for all $-1 \leq x\leq 1$, and $\,\mathrm{f}\left(\, x\,\right) > 0$ for at least three distinct points $x_{1}$, $x_{2}$, and $x_{3}$.
Note that it is straightforward to show that if the distribution function is uniform and restricted to only three points, $x_{1}$, $x_{2}$, and $x_{3}$, then $$ S = \frac{\left(\, x_{1} - x_{2}\,\right)^{\, 2} \,\left(\, x_{2} - x_{3}\right)^{\, 2} \,\left(\, x_{3} - x_{1}\,\right)^{\, 2}}{27} > 0. $$