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

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.
