Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$

Question

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!

Answer 1

Rewrite the second equation $xy+yz+zx = 1$ as $z=\frac{1-xy}{x+y}$ and evaluate

$$z+\frac1z= \frac{(x^2+1)(y^2+1)}{(x+y)(1-xy)}$$ Substitute above $z+\frac1z = \frac{z^2+1}z$ into the first equation

$$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y = \frac{5(z^2+1)}z \tag1$$ to get \begin{align} &(x^2+1)\left(\frac3{5x}- \frac{y^2+1}{(x+y)(1-xy)}\right)=0\\ &(y^2+1)\left(\frac4{5y}- \frac{x^2+1}{(x+y)(1-xy)}\right)=0\\ \end{align} which reduce to \begin{align} &3x^2y+8xy^2+2x-3y=0\tag2\\ &9x^2y+4xy^2-4x+y=0\tag3\\ \end{align} Note that 3$\times$(2) -(3) and 2$\times$(3) -(2) simplify the equations to

\begin{align} &2xy^2+x -y=0\tag4\\ &3x^2y-2x+y=0\tag5 \end{align} and, furthermore, with 3$x\times$(4) -2$y\times$(5)

$$3x^2+xy-2y^2=(3x-2y)(x+y)=0$$ Substitute the resulting $y=\frac32 x$ and $y=-x$ into (1) to obtain the real solutions $$(x,y,z)=\pm \left(\frac13,\frac12,1\right)$$

Answer 2

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