Let $X$ be a r.v and let $f \geq 0 $ be a nonincreasing function, $g$ be a nondecreasingreal-valued function. Suppose $h\geq 0$ is a function such that $h(X)$ has finite expectation with $E[h(X)f(X)] \leq E[h(X)]$. Assuming all expectations is finite, prove that
$$E[f(X)g(X)h(X)] \leq E[h(X)g(X)]. $$
What I know: If $Y \geq 0$ then we have
$$ E[Y] E[Yf(X)g(X)] \leq E[Yg(X)] E[Yf(X)],$$
Clearly, if $g(X) \geq 0$, this is immediate since we can take $Y = h(X)$. However, if this is not the case then $E[h(X)f(X)]E[h(X)g(X)] \leq E[h(X)]E[h(X)g(X)] $ may not be true anymore. I couldn't escape this...