Jtdylktuy

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

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 $L(tx)=t^8L(x)<t^4R(x)=R(tx)$.

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}
0leint_0^1Big(sum_{i=0}^2 a_i, x(t)^iBig)^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_4ge 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.