What are the solutions of this system of equations where $x,y,z \in \mathbb{R}$?
$\begin{cases} 5(x+\frac{1}{x})=12(y+\frac{1}{y})=13(z+\frac{1}{z})\\ xy+yz+zx=1 \end{cases}$
First I rewrote the equations as
$\begin{cases} 5(\frac{x^2+1}{x})=12(\frac{y^2+1}{y})=13(\frac{x^2+1}{z})\\ xy+yz+zx=1 \end{cases}$
then by multiplying the first equation with $xyz$ I got
$\begin{cases} 5yz(x^2+1)=12xz(y^2+1)=13xy(z^2+1)\\ xy+yz+zx=1 \end{cases}$
Any idea on how to proceed is appreciated.
On
Note that $x,y,z$ are of the same sign and that if $(a,b,c)$ is a solution, so is $(-a,-b,-c)$. Hence WLOG we may assume that $x=\tan \dfrac{A}{2}, y = \tan \dfrac{B}{2}, z = \tan \dfrac{C}{2}$ where $A,B,C$ are angles of a triangle
Given condition translates to $\dfrac{5}{\sin A} = \dfrac{12}{\sin B} = \dfrac{13}{\sin C} $ from which $z=1$ is immediate
Hint:
If $x<0,$ so are $y,z$ if $(p,q,r)$ is a solution, so is $(-p,-q,-r)$
WLOG $x,y,z>0;$
let $x=\cot A$ etc. to use Proving $\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1$ with $A+B+C=\pi$
Now $x+\dfrac1x=\dfrac2{\sin2A}$
If $5\cdot\dfrac2{\sin2A}=\cdots=k\implies\sin2A=\dfrac{10}k$
$$\implies\sin^22B=\sin^22C-\sin^22A=\sin2(C+A)\sin2(C-A)$$
$$\iff-\sin2B\sin2(C+A)=-2\sin2B\sin2(C-A)$$
either $\sin2B=0$ but $y+\dfrac1y>0$
or $0=\sin2(C+A)-\sin(C-A)=2\sin2B\cos2C\implies\cos2C=0\implies\sin2C=?$
Can you take it from here?