<s>[INST] Advanced algebra-precalculus Find the value of $\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}$ when $\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}$ = 1, where $w,x,y,z \in \mathbb R $ .
I have tried setting one variable equal to zero, two variables equal to zero, and many other combinations to no avail of mine. I am training for a math olympiad, and this question has been boggling my head. A solution to this would be appreciated, but not as much as resources I can use to find a definitive answer to this problem.
 [/INST] Let's try your method! Set $z=w=0$, then:
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}=1 \Rightarrow \\
\frac{x}{y}+\frac{y}{x}=1 \Rightarrow x^2+y^2=xy\Rightarrow x^2+y^2\ge 2|xy|>xy$$
So, there is no real solution. 
However, for complex numbers:
$$\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\
\frac{x^2}{y}+\frac{y^2}{x}=\frac{x^3+y^3}{xy}=\frac{(x+y)(x^2+y^2-xy)}{xy}=0.$$
Alternatively, lucky combination $w+z=0,x+y=-y$:
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}=1 \Rightarrow \\
\frac{-z}{z-y}+\frac{-2y}{y}+\frac{y}{-2y}+\frac{z}{-z-y}=1 \Rightarrow \\
\frac{z}{y-z}-\frac{z}{y+z}=\frac72 \Rightarrow \\
11z^2=7y^2\Rightarrow \\
z=\pm \sqrt{\frac{7}{11}}y$$
So:
$$y=1,x=-2,z=\sqrt{\frac7{11}}=-w\\
\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\
\frac{z^2}{z-1}+\frac{4}{1}+\frac{1}{-2}+\frac{z^2}{-z-1}=\\
\frac{2z^2}{z^2-1}+\frac72=\\
 \frac{14}{11}\cdot \left(-\frac{11}{4}\right)+\frac72=0.$$
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