Find all $x,y,z \in \mathbb R$ Such that $\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y=3 $ Find $x,y,z \in \mathbb R$ Such that :
$$\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y =3$$
My Attempt:
I’ve turned this into a system of equations :
$$\cases{\frac{1}{x}+y+z =3 \\ \frac{1}{y}+x+z =3 \\ \frac{1}{z}+x+y=3 } \iff \cases{1+xy+zx=3x \\1+xy+zy=3y \\1+xz+yz=3z}$$
Multiplying some equations by $-1$ and adding them together we get:
$$z(y-x)=3(y-x)\iff z=3$$
Notice that $y\ne x$.
You can play a little bit with $x,y$ ’s values, you will end up with :
$$(x,y,z) \in \{(3,\frac{-1}{3},3), (-3, \frac{1}{3},3)\}$$
Edit: user pointed out in comments that the first triple of my solution doesn’t work in all cases, but i don’t know why.
And the values of $x,y,z$ Can swap places because of the symmetry in the equations.
My question is what would happen if $x=y$.
And there is one more thing to notice is that one of the obvious solutions is $$x=y=z=1$$
Thank you.

If $x = y$,
$$\dfrac{1}{z} + x + y = 2x + 1/z = 3 \implies z = \dfrac{1}{3-2x}$$
Hence $x + 1/y + z = x + 1/x + 1/(3-2x) = 3$. Multiply this term by $x(3-2x)$ for $x \notin \{0, 3/2\}$ and after rearranging you will get the cubic equation
$$2x^3 - 9x^2 + 10x - 3 =(x-1)(2x^2 - 7x + 3) =  0$$
$\implies x = 1$, $x = 3$ or $ x = 0.5$.
