<s>[INST] $ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \  $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \  \frac{y}{5 \ \sqrt{y^2+1}} \ = \  \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \ xyz \ $. Find $ \ x,y,z \ . $
I substituted $ \ x,y,z \  $ as $ \ \tan(\theta_1) \ , \ \tan(\theta_2) \ , \ \tan(\theta_3) \  $  , where 
$$ \theta_1 \ + \ \theta_2 \  + \ \theta_3 \ = \  180º \ \ . $$
 [/INST] This is a comment that’s too long for the usual format.
LAcarguy’s method generalizes to any values $a,b,c$ instead
of $\frac14,\frac15,\frac16$. 
If we put 
$$
\phi(a,b,c)=ab+ac-bc, \ \psi(a,b,c)=ab+ac+bc \tag{1}
$$
and
$$
G(a,b,c)=\phi(a,b,c)\phi(b,c,a)\phi(c,a,b)\psi(a,b,c) \tag{2}
$$
then LAcarguy’s method leads to the following explicit formulas for the
solutions :
$$
x=\frac{\varepsilon}{\phi(a^2,b^2,c^2)}\sqrt{G(a,b,c)},\
y=\frac{\varepsilon}{\phi(b^2,c^2,a^2)}\sqrt{G(a,b,c)},\ 
z=\frac{\varepsilon}{\phi(c^2,a^2,b^2)}\sqrt{G(a,b,c)} \tag{3}
$$
where $\varepsilon$ is $+1$ or $-1$. 
In your initial question where $a=\frac14,b=\frac15,c=\frac16$, one finds 
$G=\frac{7}{921600}$,
$$
x=\varepsilon\frac{\sqrt{7}}{3}, \
y=\varepsilon\frac{5\sqrt{7}}{9}, \
z=3\varepsilon\sqrt{7} \tag{4}
$$
</s>