With these propositions, the trick is usually to write $x=k+e$ and $y=n+f$ where $n$ and $k$ are integers and $0\leq e,g <1$. Then your left side is $2k+\lfloor 2e\rfloor + 2n + \lfloor 2f \rfloor$. And your right side is $k+n+(k+n)+\lfloor e+f \rfloor.$ Cancelling, you need to show now that $\lfloor 2e\rfloor +\lfloor 2f \rfloor \geq \lfloor e+f \rfloor.$ You can do this by considering 4 cases, depending on whether $e$ and $f$ are less than or as big as $1/2$. E.g. if $0\leq e <1/2$ and $1/2\leq f <1$ then the left side is at least 1 and the right side is at most 1.