Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$ Find all $x,y,z>0$ such that
$$3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$$ $$xy+yz+zx = 1$$
The only solution should be $x=\frac{1}{3}$, $y = \frac{1}{2}$, $z=1$.
There is a way to do it with $x = \tan \alpha$, etc., but I would like to find an even more elementary way.
So far, I have written
$$x^2 - \frac{k}{3}x + 1 = y^2 - \frac{k}{4}y + 1 = z^2 - \frac{k}{5}z + 1 = 0$$
and noticed that at least two of $x$, $y$, $z$ must be smaller than $1$. (So in the bad cases I can express uniquely $x,y,z$.) But then?
Any help appreciated!

Hint
Note that $xy+yz+zx=1$, then we have:
$$\frac{4}{3}=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}=\frac{x+\frac{xy+yz+zx}{x}}{y+\frac{xy+yz+zx}{y}} = \frac{\frac{(x+y)(x+z)}{x}}{\frac{(y+z)(y+x)}{y}} = \frac{y(x+z)}{x(z+y)}=\frac{yz+yx}{xy+xz}$$
Do similiarly with $\frac{x+\frac{1}{x}}{z+\frac{1}{z}}$, you can imply the result
