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Let $x=\tan A$ , $y=\tan B$ and $z = \tan C$ . Then, $x+y+z=xyz$ $\implies$ $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ . This means that $A$ , $B$ and $C$ are the angles of a triangle. Then, we have:

$\begin{aligned} S & = \sum_{cyc} \frac x{\sqrt{1+x^2}} = \sum_{cyc} \frac {\tan A}{\sqrt{1+\tan^2 A}} = \sum_{cyc} \frac {\tan A}{\sqrt{\sec^2 A}} \\ & = \sum_{cyc} \frac {\tan A}{\sec A} = \sum_{cyc} \frac {\sin A}{\cos A}\cdot \cos A = \sum_{cyc} \sin A \end{aligned}$

$\begin{aligned} \implies S & = \sin A + \sin B + \sin C \\ & = \sin A + \sin B + \sin (180^\circ - (A+B)) \\ & = \sin A + \sin B + \sin (A+B) \end{aligned}$

For a fixed $A$ , the $S$ is maximum when:

$\begin{aligned} \frac {\partial S}{\partial B} & = 0 \\ \dfrac \partial{\partial B} (\sin A + \sin B + \sin (A+B)) & = 0 \\ \cos B + \cos (A+B) & = 0 \\ \cos B & = - \cos (A+B) = \cos (180^\circ - A - B) \\ \implies B & = 180^\circ - A - B \\ \implies B & = 90^\circ - \frac A2 \\ \implies S_{max}(A) & = \sin A + \sin \left(90^\circ - \frac A2 \right) + \sin \left(A + 90^\circ - \frac A2 \right) \\ & = \sin A + 2 \cos \frac A2 \end{aligned}$

And $S$ is maximum when:

$\begin{aligned} \frac {\partial S_{max}(A)}{\partial A} & = 0 \\ \frac \partial{\partial A} \left(\sin A + 2 \cos \frac A2 \right) & = 0 \\ \cos A - \sin \frac A2 & = 0 \\ \cos A & = \sin \frac A2 \\ \implies A & = 60^\circ \\ \implies B & = 90^\circ - \frac {60^\circ}2 = 60^\circ \\ \implies C & = 60^\circ \end{aligned}$

$\implies S_{max} = 3\sin 60^\circ = \dfrac {3\sqrt 3}2$

$\implies a+b+c = 3+3+2 = \boxed{8}$

Let $x=tan(A) , y=tan(B) , z=tan(C)$ .Then given condition reduces to $tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C)$ .

This is true if A,B,C are angles of a triangle (or more generally, their sum is an integral multiple of $\pi$ .)

So we need to maximize $sin(A)+sin(B)+sin(C)$ which turns out to be if all angles are $\frac{\pi}{3}$ or the triangle is equilateral.

Thus the maximum value is $\frac{3\sqrt{3}}{2}$ which means $a+b+c=8$