Can we prove this inequality holds form some region , if $(x-a)^2+(y-b)^2>R^2$ then $$ \begin{align} \ -x\arctan(x)+0.5x-y\arctan(y)+0.4y<0\\ \end{align} $$ As I can see from the WolframAlpha picture there is some region that is similar to $(x-a)^2+(y-b)^2>R^2$ but I have no idea how we can prove that and find $R$
The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $R\simeq 1/3$. We can bound $R$ more precisely looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)\in D$ then
$$-x\arctan(x)+0.5x-y\arctan(y)+0.4y\ge 0.$$
We use a fact that for each non-negative $t$ we have $\operatorname{arctan} t\le t$. Therefore we have $-t\operatorname{arctan} t\ge –t^2$ for each real $t$. Then
$$-x\arctan(x)+0.5x-y\arctan(y)+0.4y\ge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2\ge a^2+b^2-R^2.$$
So $R=\sqrt{a^2+b^2}=\sqrt{0.1025}=0.3201\dots.$ fits.
I thank to user8053696 for providing the picture:
But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.