If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$

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?

• I also tried substitution via $x=a-b, y=b-c, z=c-a$, but it got a bit horrendous... Commented Apr 14, 2016 at 7:01
• This might interest you math.stackexchange.com/questions/427222/…
– Ovi
Commented Apr 14, 2016 at 7:06

HINT:

$$2x^2+yz=2(y+z)^2+yz=(2y+z)(y+2z)$$

Now $2y+z=x+y+z+y-x=y-x$

• @MXYMXY, $$\dfrac{x^2}{2x^2+yz}=-\dfrac{x^2}{(x-y)(z-x)}$$ $$\sum\dfrac{x^2}{2x^2+yz}=\dfrac{\sum x^2(y-z)}{(x-y)(y-z)(z-x)}$$ Expand $(x-y)(y-z)(z-x)$ Commented Apr 14, 2016 at 7:09
• OK, I do feel a bit stupid. Thanks! Commented Apr 14, 2016 at 7:10

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 $$\frac{x^2}{(x-z)(x-y)}+\frac{y^2}{(y-z)(y-x)}+\frac{z^2}{(z-y)(z-x)}=...=1$$