If $x + y + z = 0$ then $x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right) = -3$

Question

Im asked to prove given $x + y + z = 0$, that: $$x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) = -3$$

Im stuck, I've tried to use that: $$x=-y-z$$ $$y=-x-z$$ $$z = -y-x$$ and trying to expand after that. But I don't know if that is a good way of proving something like this?

Answer 1

plugging $$z=-x-y$$ in the given term we obtain $$x \left( {y}^{-1}+ \left( -x-y \right) ^{-1} \right) +y \left( \left( -x-y \right) ^{-1}+{x}^{-1} \right) + \left( -x-y \right) \left( {x}^{-1}+{y}^{-1} \right) $$ simplifying this we get $$-3$$

Answer 2

$ \begin{align*} x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y}) &= \frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y} \\ &= \frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x} \\ &= \frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x} \\ &= -\frac{y}{y}-\frac{z}{z}-\frac{x}{x} \\ &= -3 \end{align*} $

Answer 3

There is an additional condition to $x+y+z=0$, namely that $x,y,z$ are all nonzero. This is implicit in what you're required to show, but it really should be stated explicitly.

Given that, you can divide $x+y+z=0$ by $x, y$ and $z$ in turn and rearrange to get:

$\frac yx + \frac zx = -1$

$\frac yz + \frac xz = -1$

$\frac xy + \frac zy = -1$

Adding and grouping terms by the respective numerators will quickly give you the desired form.

I added this answer because trying to mold something from the initial condition to the final form is a very important tip to learn well.