$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$ 
Is this claim true?
$$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$$
If so, how to prove?

I was led to this while trying to figure out the Addition Formula for Arctangent. I've looked at many questions and answers about that Formula but don't seem to have come across a proof of the above claim.
 A: 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. 
A: Below, I use these two results:


*

*$\tan^{-1}$ is strictly increasing.

*$$\tan^{-1}x+\tan^{-1}\frac{1}{x}=\begin{cases}
\pi/2, & \text{for }x>0,\\
-\pi/2, & \text{for }x<0.
\end{cases}$$
Note that if $x=0$, then both conditions hold. So below I'll assume $x\neq0$.

($\implies$) Suppose $xy<1$.
Case 1. $x>0$.
Since $y<1/x$ and $\tan^{-1}$ is strictly increasing, $$\tan^{-1}x+\tan^{-1}y<\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}.$$
Since $\tan^{-1}x>0$ and $\tan^{-1}y>-\pi/2$, $$\tan^{-1}x+\tan^{-1}y>-\frac{\pi}{2}.$$
Case 2. $x<0$. (Similar, omitted).

($\impliedby$) Suppose $\tan^{-1}x+\tan^{-1}y \in (-\pi/2,\pi/2)$.
Case 1. $x>0$.
$$\tan^{-1}x+\tan^{-1}y<\frac{\pi}{2}=\tan^{-1}x+\tan^{-1}\frac{1}{x}.$$
Since $\tan^{-1}$ is strictly increasing, $y<1/x$ or equivalently $xy<1$.
Case 2. $x<0$. (Similar, omitted).
A: Let $\;\alpha = \text{Arc}\tan x, \;\beta = \text{Arc}\tan y
\;\Rightarrow\; \alpha,\beta \;\in \;(-\pi/2, \pi/2).$
The assertion is that $\;xy < 1 \;\Leftrightarrow 
(\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$
$\Rightarrow$ 
Given that $\;xy < 1.$
WLOG $\;\alpha, \beta\;$ have the same sign, else it is immediate that $(\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$ 
$\alpha, \beta > 0\;$ and $\;(xy) < 1 \Rightarrow$ 
$0 < x, \;0 < y,\;$ and $\;y < (1/x) \;\Rightarrow$ 
$\tan \beta \;<\; \frac{1}{\tan \alpha} = \cot \alpha 
= \tan (\pi/2 \;-\; \alpha) \;\Rightarrow$ 
$\beta < (\pi/2 \;-\; \alpha) \Rightarrow 
(\alpha + \beta) \;\in \;(0, \pi/2).$
$\alpha, \beta < 0\;$ and $\;(xy) < 1 \Rightarrow$ 
$x < 0, \;y < 0,\;$ and $\;[$since $x < 0]\;$
$\;y > (1/x) \;\Rightarrow$ 
$\tan \beta \;>\; \frac{1}{\tan \alpha} = \cot \alpha 
= \tan (-\pi/2 \;-\; \alpha) \;\Rightarrow$ 
$\beta > (-\pi/2 \;-\; \alpha) \Rightarrow 
(\alpha + \beta) \;\in \;(-\pi/2, 0).$
$\Leftarrow$ 
Given that $\;(\alpha + \beta) \;\in \;(-\pi/2, \pi/2).$ 
If $\alpha$ and $\beta$ have different signs 
then $x$ and $y$ have different signs $\;\Rightarrow\; xy < 0 < 1.$ 
WLOG, $\alpha$ and $\beta$ have the same sign.
$0<\alpha, 0<\beta\;$ and $\;(\alpha + \beta) < \pi/2 
\;\Rightarrow$ 
$[\;0 < x, 0 < y\;$ and $\;\beta < (\pi/2 - \alpha)\;]
\;\Rightarrow$ 
$y = \tan \beta < \tan (\pi/2 - \alpha) = \cot \alpha = (1/x)
\;\Rightarrow xy < 1.$
$\alpha<0, \beta<0\;$ and $\;(\alpha + \beta) > -\pi/2 
\;\Rightarrow$ 
$[\;x < 0, y < 0\;$ and $\;\beta > (-\pi/2 - \alpha)\;]
\;\Rightarrow$ 
$y = \tan \beta > \tan (-\pi/2 - \alpha) = \cot \alpha = (1/x)
\;\Rightarrow $
[since $x < 0]\; xy < 1.$
