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, $$ {xyz\over(x+y+z)(y+z)z} + {yxz\over(y+x+z)(x+z)z} + {yzx\over(y+z+x)(z+x)x}={yz\over(y+z)z}.$$