<s>[INST] If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$ A problem in my homework had asked me:

When $x+y+z=0$, evaluate$$\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}$$

Without too much difficulty, one can see that the value should be $1$ using $(x,y,z)=(1,0,-1)$. 
I decided to use $x=-y-z$, which turned out not to be as difficult as initially thought.  However, would someone care to enlighten me to some other methods of doing this?
 [/INST] The first term is $$\frac{x^2}{2x^2-xy-y^2}=\frac{x^2}{(2x+y)(x-y)}=\frac{x^2}{(x-z)(x-y)}$$
Do the same transformation on each of the fractions and add them up.
So the whole expression is $$\frac{x^2}{(x-z)(x-y)}+\frac{y^2}{(y-z)(y-x)}+\frac{z^2}{(z-y)(z-x)}=...=1$$
</s>