<s>[INST] Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
$x^{2}+y^{2}+z^{2}=xy+yz+zx$
$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$
$(x+y+z)^{2}-2*3=3$
$(x+y+z)^{2}=9$
What should I do next?
 [/INST] By the calculations you have made, your system is actually eqivalent to $$\begin{cases}x^2+y^2+z^2=3\\(x+y+z)^2=9\end{cases}$$Which is, at the same time, equivalent to the systems $$\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases} \quad \begin{cases}x^2+y^2+z^2=3\\x+y+z=-3\end{cases}$$But you can check using the distance formula between a point and a plane (or in many other ways) that the planes $x+y+z=3$ and $x+y+z=-3$ are both tangent to the sphere $x^2+y^2+z^2=3$, so the only solutions are $x=y=z=\pm1$
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