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 $

Question

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.

Answer 1

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:

1. 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 $(x,y,z)=(\tfrac12,\tfrac12,\tfrac12)$.

2. 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 $$(x,y,z)=(3,3,-\tfrac13),\qquad (x,y,z)=(3,-\tfrac13,3),\qquad(x,y,z)=(-\tfrac13,3,3).$$

Answer 2

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$.