How to intuitively prove a bound for the result of this expression? In A Book of Abstract Algebra by Charles Pinter, 2nd edition, chapter 2, exercise A4, we are asked to show that, if $x$ and $y$ are chosen from the open interval $(-1, 1)$, then the value of
$$\frac{x + y}{xy + 1}$$
will also be in that interval. I am struggling to do this. I think I have a partial solution:

Suppose $x$ and $y$ are both positive. Then the numerator and denominator are both positive, and the numerator will be less than the denominator by the following reasoning. Begin with the quantity $x + 1$. In order to achieve the value of the numerator, we multiply the $1$ in this expression by $y$, which subtracts $1(1-y)$ from the final value. In order to achieve the value of the denominator, we multiply the $x$ by $y$, which subtracts $x(1-y)$ from the final value. More succinctly:
$$x + y = x + 1 - 1(1-y)$$
$$xy + 1 = x + 1 - x(1-y)$$
Since $x < 1$, we are subtracting a greater amount from the former expression. So, the numerator is less than the denominator. Nearly analogous reasoning applies to the case where $x$ and $y$ are both negative.

However, I'm not sure what to do about the case where $x$ and $y$ have opposite signs. I'm fairly happy with the intuitiveness of the argument laid out above. Is there any way to solve the whole problem in a similarly intuitive fashion, i.e. without resorting to separating the problem into four cases and using purely algebraic artifices to solve them?

One proof uses the addition formula for the hyperbolical tangent function:
$$\tanh(a+b)=\frac{\tanh(a)+\tanh(b)}{1+ \tanh(a)\tanh(b)}$$
where $a,b \in \mathbb{R}$.
$\tanh$ maps from $\mathbb{R}$ surjectively onto $(-1,1)$. Hence for any $x,y \in (-1,1)$ there exists $a,b \in \mathbb{R}$ with $x=\tanh(a)$ and $y=\tanh(b)$ which gives
$$\frac{x+y}{xy+1}=\frac{\tanh(a)+\tanh(b)}{1+ \tanh(a)\tanh(b)}=\tanh(a+b)\in(-1,1).$$
