# 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.

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 $$xyz=(x+y+z)(xy+yz+zx)$$

• nice solution. +1 and thank you! Commented Oct 8, 2016 at 15:08

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

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

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

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

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

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