Let $\chi(s)=\int\_{0}^{1}x(t)^{s}f(t)dt$,
where $x(t)$ and $f(t)$ are real valued continuous functions for
$t\in[0,1]$, and $f(t)\geq0$. 


Is it possible to show that


$\left(\chi(0)\chi(2)-\chi(1)^{2}\right)\left(\chi(4)\chi(2)-\chi(3)^{2}\right)-\left(\chi(3)\chi(1)-\chi(2)^{2}\right)^{2}\geq0$


Note: I believe that


$\chi(0)\chi(2)-\chi(1)^{2}\geq0$


$\chi(4)\chi(2)-\chi(3)^{2}\geq0$


follows from the Cauchy-Schwarz inequality.
(correct me if I am wrong about this).



A moment inequality
This inequality is false in general, by homogeneity considerations. Indeed, it can be rewritten as $L(x)\ge R(x)$, where $L(x):=\left(\chi(0)\chi(2)-\chi(1)^{2}\right)\left(\chi(4)\chi(2)-\chi(3)^{2}\right)$ and $R(x):=\chi(3)\chi(1)-\chi(2)^{2}$. Take any non-constant positive $x$, so that, by the Cauchy--Schwarz inequality, $R(x)>0$. Then for small enough $t>0$ we have $L(tx)=t^8L(x)<t^4R(x)=R(tx)$. 




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**Added in response to the OP fixing the typo:** The inequality is now true. Indeed, let $m\_j:=\chi(j)$. Without loss of generality, $m\_0 m\_2-m\_1^2>0$, by the Cauchy--Schwarz inequality. The matrix $M:=[m\_{i+j}]\_{i,j=0}^2$ is positive semidefinite. This follows because 
\begin{equation}
 0\le\int\_0^1\Big(\sum\_{i=0}^2 a\_i\, x(t)^i\Big)^2f(t)\,dt=\sum\_{i,j=0}^2 m\_{i+j}a\_i a\_j
\end{equation}
for all real $a\_0,a\_1,a\_2$. 


So, the determinant of $M$ is $\ge0$, which is equivalent to 
\begin{equation}
 m\_4\ge m\_4^\*:=\frac{m\_2^3-2 m\_1 m\_3 m\_2+m\_0 m\_3^2}{m\_0 m\_2-m\_1^2}.
\end{equation}
On the other hand, the difference between the left and right sides of your inequality (with $m\_j$ in place of $\chi(j)$) is increasing in $m\_4$, and the value of this difference is $0$ at $m\_4=m\_4^\*$. Thus, your inequality follows. 


