Given $\frac1{x+y+z}=\frac1x+\frac1y+\frac1z$ what can be said about $(x+y)(y+z)(x+z)$?

Question

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.

Answer 1

The expression $(x+y)(y+z)(z+x)$ is symmetric in $x,y,z$, so it can be expressed as $$ a(x+y+z)^3+b(x+y+z)(xy+yz+zx)+cxyz $$ for some $a,b,c$: this follows from the theory of symmetric polynomials.

With $x=1$, $y=0$, $z=0$ we obtain $$ a=0 $$ With $x=1$, $y=1$, $z=0$ we obtain $$ 8a+2b=2 $$ With $x=1$, $y=1$, $z=1$ we obtain $$ 27a+9b+c=8 $$ Thus $a=0$, $b=1$ and $c=-1$. So $$ (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz $$ which you can also verify by expanding the products.

The initial condition tells you that $$ xyz=(x+y+z)(xy+yz+zx) $$

Answer 2

$$\frac1{x+y+z}=\frac1x+\frac1y+\frac1z$$

$$\to \frac1{x+y+z}=\frac{yz+xz+xy}{xyz}$$

$$\to xyz=(yz+xz+xy)(x+y+z)$$

$$\to xyz=(yz+xz+xy)(x+y+z)$$

$$\to xyz=(x+y)(y+z)(x+z)-xyz$$

$$\to 0=(x+y)(y+z)(x+z)$$