Is this a known algebraic identity? In the course of analyzing a certain Markov chain, I once had to prove the following algebraic identity. 
Is there a slick or known proof?
For $n$-tuples $(x_1,x_2,\dots, x_n)$ of positive real numbers define
$$\mu(x_1,x_2,\dots, x_n)=\prod_{j=1}^n {x_j\over x_j+x_{j+1}+\cdots+x_n}.$$
Then if $x^\ast$ is another positive real, and $1\leq k\leq n+1$, then 
define $x^*_k$ to be the $(n+1)$-tuple $(x_1,x_2,\dots, x^*,\dots, x_n)$ where 
$x^*$ is in the $k$th place. The identity is  
$$\sum_{k=1}^{n+1}\ \mu(x^\ast_k)=\mu(x_1,x_2,\dots, x_n).$$ 
For example,
$$ {xyz\over(x+y+z)(y+z)z}
 + {yxz\over(y+x+z)(x+z)z}
 + {yzx\over(y+z+x)(z+x)x}={yz\over(y+z)z}.$$
 A: Here is a proof by induction on the number of elements in the tuple.
(Hopefully, I got the algebra right. In any case, it would be easy to verify).
First a Lemma.
Lemma: For $n \ge 2$ we have that $\displaystyle \mu(x_1, ..., x_n) = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, .., x_{n-2}, x_{n-1} + x_n)$
Proof: Easy Algebra.
Let $y = x^{*}$.
Base Case
For $n=1$ it is easy to see that $ \displaystyle \mu(y, x) + \mu(x, y) = \mu(x)$
Induction Step
Now suppose $n \ge 2$
We have that (I have changed your notation slightly, I start from 0 instead of 1)
$$\sum_{k=0}^{n-2} \mu(x_1, x_2, \dots , x_k, y, x_{k+1}, \dots,x_{n-1}, x_{n})$$
$$ = \sum_{k=0}^{n-2}\frac{x_{n-1}}{x_{n-1}+x_{n}} \mu(x_1, x_2, ..., x_k, y, \dots, x_{n-1} + x_{n})$$
Now it is easy to see that
$$\mu(x_1, \dots, x_{n-1}, y, x_n) + \mu(x_1, \dots, x_{n-1}, x_n, y)$$
$$  = \frac{x_{n-1}}{x_{n-1}+x_n} \mu(x_1, x_2, ..., x_{n-2}, x_{n-1} + x_n, y)$$
Thus we have that
$$ \sum_{k=0}^{n} \mu(x_{k}^{*}) $$ 
$$ = \sum_{k=0}^{n-1} \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (z_{k}^{*})$$
where
$$ z = (x_1, \dots, x_{n-2}, x_{n-1}+x_{n})$$
Thus by induction hyptothesis, this is the same as
$$ \frac{x_{n-1}}{x_{n-1}+x_{n}} \mu (x_1, \dots, x_{n-2}, x_{n-1}+x_{n}) $$
= $$ \mu(x_1, x_2, \dots, x_{n-1}, x_n)$$ 
A: Consider the following experiment: there is an interval of length $x_1 + \cdots + x_n$ divided into $n$ segments of sizes $x_1,\ldots,x_n$. We sample an infinite sequence of points from the interval, and write out the segments corresponding to the points. Then, we look at the order in which the segments are "discovered", i.e. which one was hit first, which was the next one, and so on. Then $\mu(x_1,\ldots,x_n)$ is the probability that the segments will be discovered in the order $x_1,\ldots,x_n$.
Now suppose we add an extra segment $x^* $, and repeat the same experiment. If we just forget about this extra segment, the new experiment is just like the old one. This is the same as running the new experiment and deleting $x^* $ from the order of discovery. Your identity follows.
