MHB What is the value of (x+y)(y+z)(z+x) if 1/(x+y+z) = 1/x + 1/y + 1/z?

[anemone]

Here is this week's POTW:

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If $\dfrac{1}{x+y+z}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$, find the value of $(x+y)(y+z)(z+x)$.

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[anemone]

No one answered last week's POTW. However, you can refer to the solution below for answer.

Spoiler

$\dfrac{1}{x+y+z}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\\\dfrac{1}{x+y+z}-\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\\\dfrac{1}{x+y+z}-\dfrac{xy+yz+zx}{xyz}=0\\\dfrac{xyz-(x+y+z)(xy+yz+zx)}{xyz(x+y+z)}=0---(1)$

$\begin{align*}(x+y)(y+z)(z+x)&=(x+y+z-z)(x+y+z-x)(z+y+x-y)\\&=((x+y+z)^2-x(x+y+z)-y(x+y+z)+xy)(x+y+z-z)\\&=(x+y+z)^3-(x+y+z)^3+(x+y+z)(xy+yz+zx)-xyz\\&=(x+y+z)(xy+yz+zx)-xyz---(2)\end{align*}$

Substituting (2) into (1) gives

$\dfrac{-(x+y)(y+z)(z+x)}{xyz(x+y+z)}=0$

Since $x,\,y,\,z$ and $x+y+z$ are not 0, we can conclude that $(x+y)(y+z)(z+x)=0$.