$\frac{1}{4}\geq \frac{xy}{(x+y)^2}=\frac{2-\sin(x+y)}{4\sin^2(x+y)}\geq \frac{1}{4}$ The equivalency condition is $x=y$ and $\sin(x+y)=1$ $\implies x=y=k\pi +\frac{\pi}{4}$, $k$ is a integer.