<s>[INST] 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.
 [/INST] Given such $x,y,z\in\Bbb{R}$, in particular you have
$$\frac1x+y+z=\frac1y+x+z,$$
and hence also $x-\tfrac1x=y-\tfrac1y$. By symmetry we see that
$$x-\frac1x=y-\frac1y=z-\frac1z=c,$$
for some constant $c$, and so $x$, $y$ and $z$ are all roots of
$$T^2-cT-1=0.\tag{1}$$
A quadratic polynomial has at most two real roots, so without loss of generality we have $x=y$. We distinguish two cases:

*

*If $x=y=z$, then we get the identity
$$\frac1x+x+x=3,$$
and hence $2x^2-3x+1=0$. This quadratic factors as
$$2x^2-3x+1=(2x-1)(x-1),$$
yielding the two solutions $(x,y,z)=(1,1,1)$ and $(x,y,z)=(\tfrac12,\tfrac12,\tfrac12)$.


*If $z\neq x$, then also $z\neq y$ and so $xz=yz=-1$ because $x$ and $z$ are distinct roots of the quadratic $(1)$. Then from
$$\frac1z+x+y=3,$$
we find that also
$$3z=1+xz+yz=1+(-1)+(-1)=-1,$$
which shows that $z=-\tfrac13$ and hence $x=y=3$. This yields the three solutions
$$(x,y,z)=(3,3,-\tfrac13),\qquad (x,y,z)=(3,-\tfrac13,3),\qquad(x,y,z)=(-\tfrac13,3,3).$$
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