$xy<1 \iff \text{arctan }x + \text{arctan }y \in (-\pi/2,\pi/2)$
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.
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).