Inequality involving expectation $X,Y$ are two RV taking values in $[1,+\infty)$. Do we have following inequality?
$$
E\left[\frac{Y}{X}\right]\geq\frac{E[Y]}{E[X]}
$$
 A: The first inequality (the one in the revised version, with the inequality sign reversed)) has no chance to be true in general, for a simple counterexample consider $Y=\frac12X$ with $\mathbb P(X\geqslant2)=1$.
The second inequality (the one in the revised version) has no chance to be true in general either, the case $Y=X^2$ would imply that $\mathbb E(X^2)\leqslant\mathbb E(X)^2$ and contradict Cauchy-Schwarz inequality for every nondegenerate $X$.
If the random variables $X$ and $Y$ are independent and positive, then by independence, $\mathbb E\left(\frac{Y}X\right)=\mathbb E(Y)\mathbb E\left(\frac1X\right)$ and, in full generality $\mathbb E\left(\frac1X\right)\geqslant\frac1{\mathbb E\left(X\right)}$ because the functions $x\mapsto x$ and $x\mapsto\frac1x$ are respectively nondecreasing and nonincreasing on $x\gt0$ (or by Jensen inequality since $x\mapsto\frac1x$ is also convex on $x\gt0$, but this is somewhat overkill). 
Finally, for every positive integrable independent random variables $X$ and $Y$,
$$
\mathbb E\left(\frac{Y}X\right)\geqslant\frac{\mathbb E(Y)}{\mathbb E\left(X\right)}.
$$
A: Let $Z=Y/X$. We then have $\mathrm{E}[XZ]=\mathrm{E}[X]\,\mathrm{E}[Z]+\mathrm{Cov}[X,Z]$. Divide both sides by $\mathrm{E}[X]$, and you get
$$
\frac{\mathrm{E}[Y]}{\mathrm{E}[X]}
=\frac{\mathrm{E}[XZ]}{\mathrm{E}[X]}
=\frac{\mathrm{E}[X]\,\mathrm{E}[Z]+\mathrm{Cov}[X,Z]}{\mathrm{E}[X]}
=\mathrm{E}\left[\frac{Y}{X}\right]+\frac{\mathrm{Cov}[X,Z]}{\mathrm{E}[X]}.
$$
From this we see that the inequality holds if the covariance between $X$ and $Z$ is negative, which will be the case if $X$ and $Y$ are independent. However, if the joint distribution of $(X,Y)$ is arbitrary, the joint distribution of $(X,Z)$ is arbitrary, and the covariance can be either positive or negative: if it is positive, the opposite inequality holds.

Example: If $X$ and $Y$ are independent and $X$ is not constant (i.e. $\mathrm{Var}[X]>0$), then
$$
\mathrm{Cov}[X,Z]=\mathrm{Cov}\left[X,\frac{Y}{X}\right]
=\mathrm{E}[Y]\cdot\mathrm{Cov}\left[X,\frac{1}{X}\right]<0.
$$
One general result that might be used to show this is
$$
\mathrm{Cov}[U,V]
=\mathrm{E}\big[\mathrm{Cov}[U,V|W]\big]
 +\mathrm{Cov}\big[\mathrm{E}[U|W],\mathrm{E}[V|W]\big]
$$
for any joint distribution $(U,V,W)$: e.g. $U=X$, $V=Y/X$, $W=X$.

Example: Let $X$ have any distribution with $\mathrm{Var}[X]>0$, and let $Y=X^2$. Then
$$
\mathrm{E}\left[\frac{Y}{X}\right]=\mathrm{E}[X]
<\mathrm{E}[X]+\frac{\mathrm{Var}[X]}{\mathrm{E}[X]}
=\frac{\mathrm{E}[X^2]}{\mathrm{E}[X]}
=\frac{\mathrm{E}[Y]}{\mathrm{E}[X]}
$$
which is a counterexample to the question. This is just the same example as stated in the previous answer by did.
