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 $-\frac\pi2\le \arctan x+\arctan y\le \frac\pi2, \arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$
You can do this. Replace $x = a + b$ and $y = 1 - ab$. Now we can get:
We can modify this to: $-x = -(a+b)$ and $-(y-1) = ab$.
Using Vieta's formula we can get solution for $a$ and $b$ solving this quadratic eqation:
$$t^2 - xt - y + 1 = 0$$
Now we can use this trigonometric identity:
$$\arctan (a) + \arctan (b) = \arctan \left( \frac{a+b}{1-ab} \right) = \arctan \left(\frac{x}{y}\right)$$
In my opinion this is as much as you can go.
You can't just remove the $arctan$, but you can get it value for $\arctan \frac{x}{y}$ using this formula:
$$arctan \frac{x}{y} \approx \frac{x}{y} - \frac{1}{3}\left(\frac{x}{y}\right)^3 + \frac{1}{5}\left(\frac{x}{y}\right)^5...$$
This is Taylor-Maclaurin series and you'll get closer and closer to the true value with each step, although you will not get exactly the true falue, but you'll be close enough.
The summation formula for the series is:
$$\arctan \frac{x}{y} = \sum_{i=0}^\infty (-1)^i\frac{1}{2i+1}\left(\frac{x}{y}\right)^{2i+1}$$
Best Answer
It is enough to prove that $\arctan x+\arctan y <\pi/2$ since we can change $x$ to $-x$ and $y $ to $-y$ to get the lower bound.
Let $y >0$. Now $\arctan x+\arctan y$ is a strictly increasing function of $x$ so it is enough to prove that $\arctan y+\arctan \frac 1y \leq \pi /2$. You can check that the derivative of the left side is negative for $y <1$ positive for $y>1$. Hence the minimum value is attained when $y=1$. But the value when $y=1$ is $\frac {\pi} 4+\frac {\pi} 4=\frac {\pi} 2$.
I will leave the case $y <0$ to you.