<s>[INST] System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z;   \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Rightarrow 3y(x^2+1)=4x(y^2+1).$$ Now, from the second row, we have $$3y(x^2+xy+xz+yz)=4x(y^2+xy+xz+yz); \Rightarrow (x+y)(xy+4xz-3yz)=0.$$
Thus, $$x=-y; z=\frac {xy}{3y-4x}.$$
But if we replace $y$ with $-x$, we obtain a complex solution. The same happens when we replace $z$ with $\frac {xy}{3y-4x}$ in the second row because it takes us back to the first expression $x=-y$.
Finally, in case we try to deal with $$\frac{3(x^2+1)}x=\frac{5(z^2+1)}z, z=\frac {xy}{3y-4x},$$
then it complicates the problem with huge expressions.
Wolfram Alpha shows that the real solutions are $(x;y;z)=(\frac 13;\frac 12;1)\cup(-\frac 13;-\frac 12;-1)$, but I do not know how to get to them. I would appreciate any help.
Thank you in advance.
 [/INST] (Not a completely rigorous proof, but too long for a comment.)
Let $\,x=\tan(a)\,$ and similar, then $\,\frac{x^2+1}{x}=\frac{2}{\sin(2a)}\,$, so the equations can be written as:
$$
\frac{3}{\sin(2a)}=\frac{4}{\sin(2b)}=\frac{5}{\sin(2c)} \tag{*}
$$
Also, $\,xy+yz+zx=1 \implies a+b+c=(2k+1)\frac{\pi}{2}\,$, by proving a converse of if $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $, and therefore $\,2a+2b+2c=(2k+1)\pi\,$.
For positive $\,x,y,z\,$ the relation $\,(*)\,$ can be thought of as the law of sines in a right $3-4-5$ triangle with angles $\,2a,2b,2c = \frac{\pi}{2}\,$. It follows that the following are solutions:

*

*$\sin(2a)=\frac{3}{5} \implies \frac{x^2+1}{x} = 2 \cdot \frac{5}{3} \implies x = \frac{1}{3}$


*$\sin(2b)=\frac{4}{5} \implies \frac{y^2+1}{y} = 2 \cdot \frac{5}{4} \implies y = \frac{1}{2}$


*$\sin(2c)=1 \implies \frac{z^2+1}{z} = 2 \cdot 1 \implies z = 1$
This gives the positive solution $\,\left(\frac{1}{3}, \frac{1}{2}, 1\right)\,$, and
the given relations require $\,x,y,z\,$ to all have the same sign, so there is one symmetric solution in negative numbers $\,\left(-\frac{1}{3}, -\frac{1}{2}, -1\right)\,$.
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