If $\arctan x=A,\arctan y=B;$ $\tan A=x,\tan B=y$
We know, $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
So, $$\tan(A+B)=\frac{x+y}{1-xy}$$
$$\implies\arctan\left(\frac{x+y}{1-xy}\right)=n\pi+A+B=n\pi+\arctan x+\arctan y $$ where $n$ is any integer
As the principal value of $\arctan z$ lies $\in[-\frac\pi2,\frac\pi2], -\pi\le\arctan x+\arctan y\le\pi$
$(1)$ If $\frac\pi2<\arctan x+\arctan y\le\pi, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y-\pi$ to keep $\arctan\left(\frac{x+y}{1-xy}\right)\in[-\frac\pi2,\frac\pi2]$
Observe that $\arctan x+\arctan y>\frac\pi2\implies \arctan x,\arctan y>0\implies x,y>0 $
$\implies\arctan x>\frac\pi2-\arctan y$
$\implies x>\tan\left(\frac\pi2-\arctan y\right)=\cot \arctan y=\cot\left(\text{arccot}\frac1y\right)\implies x>\frac1y\implies xy>1$
$(2)$ If $-\pi\le\arctan x+\arctan y<-\frac\pi2, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y+\pi$
Observe that $\arctan x+\arctan y<-\frac\pi2\implies \arctan x,\arctan y<0\implies x,y<0 $
Let $x=-X^2,y=-Y^2$
$\implies \arctan(-X^2)+\arctan(-Y^2)<-\frac\pi2$
$\implies \arctan(-X^2)<-\frac\pi2-\arctan(-Y^2)$
$\implies -X^2<\tan\left(-\frac\pi2-\arctan(-Y^2)\right)=\cot\arctan(-Y^2)=\cot\left(\text{arccot}\frac{-1}{Y^2}\right) $
$\implies -X^2<\frac1{-Y^2}\implies X^2>\frac1{Y^2}\implies X^2Y^2>1\implies xy>1 $
$(3)$ If $-\frac\pi2\le \arctan x+\arctan y\le \frac\pi2, \arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$
Over the given interval we have $\arctan\cot A=\frac{\pi}{2}-A$ and, by setting $t=\tan A$:
$$\begin{eqnarray*}&&\tan\left(\arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)\right)=\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}\\&=&\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}=\frac{\frac{1}{t^3}+\frac{t}{1-t^2}}{1-\frac{1}{t^2(1-t^2)}}=\frac{1-t^2+t^4}{t(t^2-t^4-1)}=-\frac{1}{t}\end{eqnarray*}$$
so:
$$ \arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)=\frac{\pi}{2}+A $$
and the sum of the three arctangents is $\color{red}{\pi}$ as wanted. Another chance is given by differentiating such a sum wrt to $A$ and check that the derivative is zero oven the given interval, so the sum equals its value in the point $A=\frac{\pi}{8}$, for instance.
Best Answer
Fix, as usual:
$$ -\frac{\pi}{2}<\gamma=\arctan(t)<\frac{\pi}{2} $$
now we have: $$ \tan (\gamma)=\tan(\alpha+\beta)=\frac{x+y}{1-xy}=t $$ and, if $xy>1$ we have the two cases ($x$ and $y$ have the same sign): $$ x>0, y>0 \rightarrow t<0 \rightarrow \gamma<0\rightarrow \alpha+\beta=\gamma+\pi $$ $$ x<0, y<0 \rightarrow t>0 \rightarrow \gamma>0\rightarrow \alpha+\beta=\gamma-\pi $$