<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] HINT:
$$2x^2+yz=2(y+z)^2+yz=(2y+z)(y+2z)$$
Now $2y+z=x+y+z+y-x=y-x$
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