Let $Y$ be a RV taking values in $[0,+\infty)$, and let $X$ be a RV taking values in $[1, +\infty)$, such that $X \ge Y$. Do we have following inequality?
$ E[Y/X]≥E[Y]/E[X] $
This question is closely related to this earlier question on ratios, but here $Y$ is allowed to be zero, and $X$ dominates $Y$ - which is why the counterexample $Y = X^2$ with $Var(X) > 0$ presented there doesn't apply here.
Still false, sorry. Let $X$ take values $1$ and $3$ with equal probability, and define $Y:=X-1$. Then $E(X)=2$, $E(Y)=1$, $E(Y/X) = 1/3$. Moreover $Y$ can take value $0$, and $X$ dominates $Y$, but $$ E(Y/X) = {1\over3} < {1\over2} = {E(Y)\over E(X)} .$$ What's the intuition here? In general the inequality that you want to be true is valid if and only if $Y/X$ is negatively correlated with $X$, since it's equivalent to $$E[(Y/X) X] \le E(Y/X) E(X) \Longleftrightarrow{\rm Cov}(Y/X,X)\le0.$$ So if you're looking for a counterexample, you want to find $Y/X$ that is positively correlated with $X$, while satisfying $Y\le X$. The choice $Y:=X-1$ will do.