If $x + y + z = 0$ then $x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right) = -3$ 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?
 A: $
\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*}
$
A: 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.
A: 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 $$-3$$
