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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.

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\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}

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Hint

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

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

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  • $\begingroup$ That means that a = 0, correct? So the whole thing is equal to zero? $\endgroup$ Jun 4 '19 at 4:46
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    $\begingroup$ @HugoBethancourt, Correct you are $\endgroup$ Jun 4 '19 at 4:59
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    $\begingroup$ Was not previously aware of the 'cyclic sum'. It is quite useful and elegant as a tool. Thank you for sharing your knowledge! $\endgroup$ Jun 4 '19 at 5:03
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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.

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    $\begingroup$ Wow. This is a vastly different approach to other answers...Thank you for sharing! $\endgroup$ Jun 4 '19 at 18:30
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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*}

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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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  • $\begingroup$ 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 $t = -1$). $\endgroup$
    – user296602
    Jun 4 '19 at 16:44
  • $\begingroup$ Yep, that's where I realized I couldn't get real solutions by setting two variables equal to zero. $\endgroup$ Jun 4 '19 at 18:24

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