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.
 A: \begin{align}
\sum_{\text{cyc}}\frac{w^2}{x+y+z} &= \sum_{\text{cyc}}\left(\frac{w^2}{x+y+z}+w\right)-\sum_{\text{cyc}}w\\
&= \sum_{\text{cyc}}\left(\frac{w(w+x+y+z)}{x+y+z}\right)-\sum_{\text{cyc}}w\\
&= \sum_{\text{cyc}}w\sum_{\text{cyc}}\left(\frac{w}{x+y+z}\right)-\sum_{\text{cyc}}w\\
&= 0
\end{align}
A: Hint
If $a=\sum_{\text{cyc}}\dfrac {w^2}{x+y+z}$, then
$a+w+x+y+z=\sum_{...}\left(w+\dfrac{w^2}{x+y+z}\right)=(w+x+y+z)\sum\dfrac w{x+y+z}=(w+x+y+z)1$
A: Let $$a=\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}$$
and let 
 $$b=\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}$$
with $w, x, y, z \in \mathbb{R} $

The question you are asked then becomes

If $b=1$, then what is the value of $a$? 


I will show something interesting. I will show that if the question is answerable with only the information you are given, then the answer must be $0$.

Notice that the value of $b$ does not change if we multiply all four of $w, x, y, z$ by any nonzero real constant $c$. 
However, if we do this, the value of $a$ does change, by a factor of $c$. 
If the question can be answered then there is just one possible value for $a$. But then $ac=a$ for all nonzero real $c$. 
The only real number that is invariate, under multiplication by all choices for the nonzero constant $c$, is zero. 

We have shown that either the question cannot be answered with only the given information, or $$a=0$$

Since you mentioned that you are preparing for a mathematics olympiad, this type of reasoning (involving an assumption that the question does have a definite answer) could be useful. 
A: Here is a way that starts with the given expression $\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}$ and works out the searched for one.
Let


*

*$S = \sum_{cyc}w\Rightarrow \frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y} = \sum_{cyc}\frac{w}{S-w} = 1$
Now, we have
\begin{eqnarray*} S
& = & S\underbrace{\sum_{cyc}\frac{w}{S-w}}_{=1} \\
& = & \sum_{cyc}\frac{(S-w+w)w}{S-w} \\
& = & \sum_{cyc}\left(w + \frac{w^2}{S-w} \right) \\
& = & S +\sum_{cyc}\frac{w^2}{S-w} \\
& \Rightarrow & \sum_{cyc}\frac{w^2}{S-w} =0
\end{eqnarray*}
A: 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:
$$y=1,x=-2,z=\sqrt{\frac7{11}}=-w\\
\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\
\frac{z^2}{z-1}+\frac{4}{1}+\frac{1}{-2}+\frac{z^2}{-z-1}=\\
\frac{2z^2}{z^2-1}+\frac72=\\
 \frac{14}{11}\cdot \left(-\frac{11}{4}\right)+\frac72=0.$$
