Let $X, Y$ be two identically distributed (i.d.) positive random variables. If they are furthermore independent, from the Cauchy-Schwarz inequality (last step), one has
$$ E[X/Y] = E[X].E[1/Y] = E[X].E[1/X] \geq E[\sqrt{X}/\sqrt{X}]^2 = 1 $$
Futhermore, if one doesn't assume independence of $X$ and $Y$ (but one still assumes that $X\sim Y$), by Jensen inequality one gets$$ \log E[X/Y] \geq E[\log X - \log Y] $$ and the right hand side is $0$ if $\log X$ is assumed integrable, and this gives again $(*)\, E[X/Y] \geq 1$
I am pretty sure (still in the case where $X$ and $Y$ are identically distributed and possibly dependent) that $(*)$ holds true without the hypothesis $\log X \in L^1$, but my attempts don't work (I tried to replace $X$ and $Y$ by truncated versions, but the dominated convergence theorem doesn't work). Maybe there is another way to do it?