<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] 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 $\left(-\frac{1}{3},-\frac{1}{2},-1\right)$
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