Given $\frac1{x+y+z}=\frac1x+\frac1y+\frac1z$ what can be said about $(x+y)(y+z)(x+z)$? 
If $\frac1{x+y+z}=\frac1x+\frac1y+\frac1z$ where $xyz(x+y+z)\ne0$, then the value of $(x+y)(y+z)(z+x)$ is
(A) zero
(B) positive
(C) negative
(D) non-negative

I substituted $x=-y$ and the equality was established. In the given expression the factor $(x+y)$ would be 0 and the result would be 0. But how should I proceed to show that 0 is the only possible result? I did some algebraic manipulations which do not seem to be of any use. I also believe that we can assume the variables can only be real – this might somehow play a role. Thanks in advance.

$$\frac1{x+y+z}=\frac1x+\frac1y+\frac1z$$
$$\to \frac1{x+y+z}=\frac{yz+xz+xy}{xyz}$$
$$\to xyz=(yz+xz+xy)(x+y+z)$$
$$\to xyz=(yz+xz+xy)(x+y+z)$$
$$\to xyz=(x+y)(y+z)(x+z)-xyz$$
$$\to 0=(x+y)(y+z)(x+z)$$
