$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is?

Question

If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?

$$A)\ 4\quad B)\ 3\quad C)\ 2\quad D)\ 1$$

My attempt:
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k>2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use.
2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.

Answer 1

Other answers have shown that the question is flawed. For fun, let's remove the condition that $x,y,z$ are positive, but keep the requirement that $x,y,z$ are distinct, and try to find all solutions $(x,y,z)$.

Trivially, $x,y,z \neq 0$. If we set $x+\dfrac{1}{y} = y+\dfrac{1}{z} = z + \dfrac{1}{x} = k$, for som real number $k$, then \begin{align} x &= k - \dfrac{1}{y} \\ y &= k - \dfrac{1}{z} \\ z &= k - \dfrac{1}{x} \end{align}

If we substitute equation (3) into (2) and then substitute that into (1), we get $$x = k-\dfrac{1}{k-\dfrac{1}{k-\tfrac{1}{x}}}.$$

By unraveling this fraction, we eventually get the equation $$(k^2-1)(x^2-kx+1) = 0.$$

If $k \neq \pm 1$, then the solutions $x$ will satisfy $$x^2-kx+1 = 0 \iff x = k - \dfrac{1}{x} = z,$$ and we won't have distinctness.

If $k = 1$, then $z = 1-\dfrac{1}{x} = \dfrac{x-1}{x}$ and $y = 1-\dfrac{1}{z} = -\dfrac{1}{x-1}$.

So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$, all of which satisfy $xyz = -1$.

If $k = -1$, then $z = -1-\dfrac{1}{x} = -\dfrac{x+1}{x}$ and $y = -1-\dfrac{1}{z} = -\dfrac{1}{x+1}$.

So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$, all of which satisfy $xyz = 1$.

Therefore, the only solutions where $x,y,z$ are distinct are all of the form $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$ for $x \neq 0,1$ or $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$ for $x \neq 0,-1$. (It's easy to check that these satisfy the distinctness condition.) Also, the only possible values of $xyz$ are $\pm 1$.

Answer 2

From the conditions we have: $$x-y=\frac{y-z}{yz}$$ $$z-x=\frac{x-y}{xy}$$ and $$y-z=\frac{z-x}{xz}$$ Thus, $xyz=1$, but why it holds?

Answer 3

By multiplying the first equation by $yz$, the second by $xz$, and the equation connecting the first and third expression by $xy$, we get

$$xyz+z = y^2z+y$$ $$xyz+x = z^2x+z$$ $$xyz+y = x^2y+x$$

Adding these three, cancelling $x+y+z$ from both sides, and dividing by $xyz$ gives $$\frac{x}{z}+\frac{y}{x}+\frac{z}{y} = 3$$

But by AM-GM, the LHS is $\ge 3$ and so we must have equality in AM-GM, meaning $\frac{x}{z} = \frac{y}{x} = \frac{z}{y}$, and so $x^2 = yz, y^2 = xz, z^2 = xy$. Dividing the first by the second gives $\frac{x^2}{y^2} = \frac{y}{x}$ and so $x^3 = y^3 \implies x=y$. Similarly, $y=z$. This contradicts that $x,y,z$ are distinct, so the question is flawed.

Answer 4

Since there is a question, there is an answer.

By inspection, $x=y=z$ satisfies the equations.

Therefore $xyz = x^3$.

But the only cube listed is $1$, so that must be the answer.

Meta-reasoning strikes again!