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.

\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}

Hint

If $$a=\sum_{\text{cyc}}\dfrac {w^2}{x+y+z}$$, then

$$a+w+x+y+z=\sum_{...}\left(w+\dfrac{w^2}{x+y+z}\right)=(w+x+y+z)\sum\dfrac w{x+y+z}=(w+x+y+z)1$$

• That means that a = 0, correct? So the whole thing is equal to zero? Jun 4, 2019 at 4:46
• @HugoBethancourt, Correct you are Jun 4, 2019 at 4:59
• Was not previously aware of the 'cyclic sum'. It is quite useful and elegant as a tool. Thank you for sharing your knowledge! Jun 4, 2019 at 5:03

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.

• Wow. This is a vastly different approach to other answers...Thank you for sharing! Jun 4, 2019 at 18:30

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 $$\begin{eqnarray*} S & = & S\underbrace{\sum_{cyc}\frac{w}{S-w}}_{=1} \\ & = & \sum_{cyc}\frac{(S-w+w)w}{S-w} \\ & = & \sum_{cyc}\left(w + \frac{w^2}{S-w} \right) \\ & = & S +\sum_{cyc}\frac{w^2}{S-w} \\ & \Rightarrow & \sum_{cyc}\frac{w^2}{S-w} =0 \end{eqnarray*}$$

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: $$y=1,x=-2,z=\sqrt{\frac7{11}}=-w\\ \frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\ \frac{z^2}{z-1}+\frac{4}{1}+\frac{1}{-2}+\frac{z^2}{-z-1}=\\ \frac{2z^2}{z^2-1}+\frac72=\\ \frac{14}{11}\cdot \left(-\frac{11}{4}\right)+\frac72=0.$$

• I think that this answer is pretty problematic: note that $x/y + y/x = 1$ admits no real solutions because $t + 1/t$ is minimized at $t = 1$ (or maximized at $t = -1$).
– user296602
Jun 4, 2019 at 16:44
• Yep, that's where I realized I couldn't get real solutions by setting two variables equal to zero. Jun 4, 2019 at 18:24