If $\sqrt{x+y}+\sqrt{y+z}=\sqrt{x+z}$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=?$

Question

If $\sqrt{x+y}+\sqrt{y+z}=\sqrt{x+z}$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=?$

I really am stumped on this problem. I squared the first equation and found that

$-y = \sqrt{(x+y)(y+z)}$. So this means that $(x+y)(y+z) = y^2$, so $(x+y)(y+z)-y^2=0$.

I then expanded $(x+y)(y+z)$ using FOIL to make $yx+y^2+yz+xz-y^2=0$.

I simplified the previous equation to read $yx+yz+xz=0$

Then I isolated $x$ and $z$ from the above equation.

$y = -\sqrt{(x+y)(y+z)}=-\frac{x(y+z)}{z}, x=-\frac{y(x+z)}{z},z=-\frac{y(x+z)}{x}$ are all of my deductions, but I don't know how this helps me at all.

I also entertained the notion that you don't need the unknowns to work out $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$. It was a multiple choice answer, with the answers being $-1,0,1,$ also with $xyz$ and $x+y+z$ being possible answers. Any help on solving this problem is appreciated.

Answer 1

$\displaystyle -y = \sqrt{(x+y)(y+z)}\implies(x+y)(y+z) = (-y)^2=y^2$

$\displaystyle\implies y^2=(x+y)(y+z)=xy+y^2+xz+yz\iff xy+yz+zx=0 $

If $\displaystyle x=0,yz=0\implies$ at least one of $y,z$ is $0$

If $y=0,$ from the given relation becomes an identity

Assuming $xyz\ne0,$ divide either sides by $xyz$

Answer 2

Squaring both sides of $-y = \sqrt{(x+y)(y+z)}$ should give you $y^2 = (x+y)(y+z)$.