Is this statement true: $\mathsf{Cov}(X,Y)\geq 0$, $\mathsf{P}(Y>0)=1$, show that $\mathsf{Cov}\Big(X,\dfrac{1}{Y}\Big)\leq 0$

Question

I was trying to prove a problem and I got stuck at a point. The problem leads to a point where I have to show:

Let, $X$ and $Y$ are two r.v. If $\mathsf{Cov}(X,Y)\geq 0$ and $\mathsf{P}(Y>0)=1$, then show that $\mathsf{Cov}\Big(X,\dfrac{1}{Y}\Big)\leq 0$.

But I could not find a way to go from $Y$ to $\dfrac{1}{Y}$, also I have a feeling that there may be a counterexample of the problem. Any help would be appreciated.

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Thanks to Einar Rødland, but in the problem I did not have $(X,Y)$ and $(X, 1/Y)$ same in distribution.

The problem was to show if $\mathsf{E}\Big(\dfrac{Z_1-Z_2}{Z_1+Z_2}\Big)\geq 0$, then $\mathsf{E}(Z_1-Z_2)\geq 0$, where $Z_1$ and $Z_2$ are different positive r.v. and $\mathsf{Cov}(Z_1+Z_2,Z_1-Z_2)\geq 0$

Answer 1

Let $\Pr(X=-1, Y=1)=1/2$ and $\Pr(X=1, Y=2)=\Pr(X=1, Y=1/2)=1/4$.

Note that $(X,Y)$ and $(X,1/Y)$ have the same distribution, both with $\text{E}[X]=0$, which made the example a little easier to make and explain.

That makes $\text{Cov}[X,Y]=\text{Cov}[X,1/Y]=1/8$.