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.
 A: (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 $\,\left(-\frac{1}{3}, -\frac{1}{2}, -1\right)\,$.
A: Expanding on OP's work and my comments:
From the first double equality it is obvious that $x,y,z\ne 0$, and $x, y$ and $z$ have the same sign (since the numerators are positive). So one also has $x+y,y+z$, and $z+x\ne 0$
Then as in OP's idea, using the given:
$$xy+yz+xz=1 \tag{1}$$
One gets:
$$x^2+1=x^2+xy+yz+xz=x(x+y)+z(x+y)=(x+y)(x+z)$$
So the first double equality can be rewritten as:
$$\frac{3(x+y)(x+z)}{x}=\frac{4(y+x)(y+z)}{y}=\frac{5(z+x)(z+y)}{z}$$
From the first and middle side of the double equality above, as $x+y\ne 0$ one gets:
$$\frac{3(x+z)}{x}=\frac{4(y+z)}{y}\text{ which can be rewritten as }\frac{3z}{x}=\frac{y+4z}{y}$$
Which gives
$$xy-3yz+4xz=0\tag{2}$$
Similarly, from the middle and right side of the double inequality one gets:
$$5xy+yz-4xz=0\tag{3}$$
From $(1)$, $(2)$ and $(3)$ by denoting $u=xy, v=yz, w=zx$, one gets the linear system:
$$\left\{
\begin{alignat}{4}
u&+&v&+&w&=1\\
u&-&3v&+&4w&=0\\
5u&+&v&-&4w&=0
\end{alignat}
\right.$$
By solving with one's favorite method, one gets $u=\frac{1}{6}, v=\frac{1}{2}, w=\frac{1}{3}$
Furthermore $(xyz)^2=uvw=\frac{1}{36}$, so $xyz=\pm\frac{1}{6}$
So $x=\frac{xyz}{v}=\pm\frac{1}{3}, y=\frac{xyz}{w}=\pm\frac{1}{2}, z=\frac{xyz}{u}=\pm 1$
As $x,y$ and $z$ have the same sign, one gets the solutions $(x,y,z)=\left(\frac{1}{3},\frac{1}{2},1\right)$ and $\left(-\frac{1}{3},-\frac{1}{2},-1\right)$
A: HINT.-Equalizing to $t$ each of the expressions in first line we have
$$3(x^2+1)=tx\\4(y^2+1)=ty\\5(z^2+1)=tz$$ we deduce
$$x=\frac{t\pm\sqrt{t^2-36}}{6}\\y=\frac{t\pm\sqrt{t^2-64}}{8}\\z=\frac{t\pm\sqrt{t^2-100}}{10}$$Now taking into account the second line of the proposed system we have the solution given by Wolfram:
$$\left(\frac{t\pm\sqrt{t^2-36}}{6}\right)\left(\frac{t\pm\sqrt{t^2-64}}{8}\right)+\left(\frac{t\pm\sqrt{t^2-36}}{6}+\frac{t\pm\sqrt{t^2-64}}{8}\right)\left(\frac{t\pm\sqrt{t^2-100}}{10}\right)=1$$ but it is not necessary if we before that $t=10$ is a good value.
