If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$? 
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?

The answer says $1$, but I am not sure how to derive it.

From the fact that Expressions $1$ and $2$ are equal, we obtain
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}.$$
From the fact that Expressions $2$ and $3$ are equal, we obtain
$$b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}.$$
From the fact that Expressions $3$ and $1$ are equal, we obtain
$$c-a=\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}.$$
Multiply the left-hand sides, the right-hand sides.  We get
$$(a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)\ne 0$.  We conclude that 
$(abc)^2=1$.  This yields the two possibilities $abc=1$ and $abc=-1$.
In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=\pm 1$.
But it would be nice to show that there are solutions of the desired type.  So let's find such a solution, with $abc=1$.
Look for a solution with $a=1$.  Then we need $c=1/b$.  In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$.  We conclude that 
$$a=1,\quad b=-\frac{1}{2},\quad c=-2$$
is a solution of the desired type, with $abc=1$.  By changing all the signs, we find that 
$$a=-1,\quad b=\frac{1}{2},\quad c=2$$ 
is a solution with $abc=-1$.
Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $t\ne 0$.  Now assume that $abc=\pm 1$.  We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.  
Let $a=t$.  Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!
