The OP has just been edited from an apparently absolutely continuous version to a discrete one. So, I gather that the integrals are now to be understood with respect to the counting measure, that is, as the sums.

However, there is no essential difference here between the absolutely continuous and discrete versions, and counterexamples for the two versions can be quite similar. For instance, suppose that $x_1$ and $x_2$ are random variables taking only nonnegative integral values such that
\begin{equation}
F_1(x)=1-\frac1{\ln x}\quad\text{and}\quad f_2(x)=\frac C{x\ln^2 x}
\end{equation}
for integers $x\ge3$ and $F_1(x)=f_2(x)=0$ for integers $x<3$, where $C$ is a positive real constant such that $f_2$ is a pmf.
Then
\begin{multline}
\int_{0}^{\infty}x{f_2(x)}(1-F_1(x))=\sum_{0}^{\infty}x{f_2(x)}(1-F_1(x))
=\sum_{3}^{\infty}\frac C{\ln^3 x}=\infty \\
>c\ge c\int_{c}^{\infty}f_1(x)
\ge c\int_{c}^{\infty}f_1(x)F_2(x)
\end{multline}
for any real $c>0$, so that we have the opposite of the desired inequality.