If you are interested in the best approximation of $\arctan(x)$ by the function of the form $$\frac {ax} {b+\sqrt{c+dx^2}},$$ then the Maple global optimizer DirectSearch does the job in such way: the code
$$A := proc (alpha, beta, delta, eps) DirectSearch:-GlobalSearch(abs(alpha*x/(beta+sqrt(delta+eps*x^2))-arctan(x)), {x = -infinity .. infinity}, solutions = 1) end proc: $$ $$ DirectSearch:-Search(proc (a, b, c, d) -> A(a, b, c, d)[1, 1] , {abs(a) <= 20, abs(b) <= 20, abs(c) <= 20, abs(d) <= 20});$$ produces
$$ [ 0.0, \left[ \begin {array}{c} 1.76051623816363034
\\ - 0.155583793604506360\\
1.72404758813357750\\ 1.36875140919455384
\end {array} \right] ,87]
$$
The following formula is well known and you can find it here.
$$\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)$$
(of course up to a multiple of $\pi$ with $ab\neq 1$). Then
$$\arctan(a+b) + \arctan(a-b) = \arctan\left(\frac{2 a}{1-a^2+b^2}\right)$$
$$\arctan(a+b) - \arctan(a-b) = \arctan\left(\frac{2 b}{1+a^2-b^2}\right)$$
Hence we have
\begin{align*}
2 \arctan(a+b) &= \arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\arctan\left(\frac{2 b}{1+a^2-b^2}\right)\\[2mm]
\arctan(a+b) &= \frac{1}{2}\arctan\left(\frac{2 a}{1-a^2+b^2}\right)+\frac{1}{2}\arctan\left(\frac{2 b}{1+a^2-b^2}\right)
\end{align*}
Best Answer
It is sufficient that the sum of the three arctangents is between $\pm\pi/2.$
For $x,y,z\in\mathbb R,$ the value of $$\frac{x+y+z-xyz}{1-xy-xz-yz}\in\mathbb R\cup\{\infty\}$$ is always equal to $$\tan(\arctan x + \arctan y + \arctan z)\in\mathbb R\cup\{\infty\}, $$ since $$\tan(\arctan x) = x \text{ for every } x\in\mathbb R$$ and for all $a,b,c\in\mathbb R,$ $$ \tan(a+b+c) = \frac{\tan a+\tan b+\tan c-\tan a\tan b\tan c}{1 - \tan a\tan b - \tan a \tan c - \tan b\tan c} \in\mathbb R\cup\{\infty\}. $$ (This $\text{“}\infty\text{”}$ is neither $+\infty$ nor $-\infty,$ but is approached by going in either the positive or the negative direction, so $\mathbb R\cup\{\infty\}$ is topologically a circle.)
However, in some cases $\arctan (\tan a) \ne a,$ namely in those cases in which $a\notin(-\pi/2,+\pi/2).$