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My approach is naive: Given $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,

$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$

What to do next?

Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.

Is there any trick to such genre of problems?

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  • $\begingroup$ Are there four equations? $\endgroup$ Apr 4, 2018 at 19:23
  • $\begingroup$ No three only. I have quoted the exact question here. $\endgroup$
    – Saradamani
    Apr 4, 2018 at 19:24
  • $\begingroup$ From where do you got this system? $\endgroup$ Apr 4, 2018 at 19:45

4 Answers 4

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Given $\;x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}\,$,

$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+\color{red}{3}$

That's actually a good approach, and it does in fact work out nicely, but:

  • in order to reverse the fractions, you have to assume $\,x,y,z \ne 0\,$; however $\,x=y=z=0\,$ is a solution, which you lose if you don't state that assumption upfront;

  • you made a mistake in the calculations, the second line should rather be:

$$ \begin{align} &\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2} \cdot \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\color{red}{\frac{3}{2}} \\[5px] \iff\quad &\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} - 2 \cdot \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) + 3 = 0 \\[5px] \iff\quad &\left(\frac{1}{x}-1\right)^2 + \left(\frac{1}{y}-1\right)^2 + \left(\frac{1}{z}-1\right)^2 = 0 \\[5px] \iff\quad &x=y=z=1 \end{align} $$

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    $\begingroup$ Ahh such a smooth one.I loved the way you showed it.A trillion thanks. $\endgroup$
    – Saradamani
    Apr 5, 2018 at 4:08
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We have $x\ge0$, $y\ge0$ and $z\ge0$.

One possible solution is $x=y=z=0$.

If one of $x,y,z$ is non-zero, the other two are also non-zero.

Now suppose that all of them are non-zero.

$$xyz=\frac{8x^2y^2z^2}{(1+x^2)(1+y^2)(1+z^2)}\le\frac{8x^2y^2z^2}{(2x)(2y)(2z)}=xyz$$

The equality holds if and only if $x^2=y^2=z^2=1$ and this would imply that $x=y=z=1$.

There are totally two solutions.

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It is easy to show that $$x,y,z>0$$ Clearly that $$x=y=z=0$$ is one solution. Now we assume $$x\geq y\geq z$$ this is equivalent to $$\frac{z^2}{1+z^2}\geq \frac{2x^2}{1+x^2}\geq \frac{2y^2}{1+y^2}$$ From here we get $$\frac{z^2}{1+z^2}\geq \frac{x^2}{1+x^2}$$ this implies $$z\geq x$$ and so on, and we get $$x=y=z$$ and $$1+x^2=2x$$ this means $$(x-1)^2=0$$ or $$x=y=z=1$$

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  • $\begingroup$ Simplest and easiest proof thanks Dr Graubner $\endgroup$
    – Saradamani
    Apr 4, 2018 at 19:37
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From the system of equations, you can infer that $x,y,z \ge 0$ and consider the function $f(a) = \dfrac{2a^2}{1+a^2}=2 - \dfrac{2}{1+a^2}$. Thus if $a > b>0 \implies a^2>b^2\implies 1+a^2>1+b^2\implies \dfrac{2}{1+a^2} < \dfrac{2}{1+b^2}\implies -\dfrac{2}{1+a^2} > -\dfrac{2}{1+b^2}\implies 2-\dfrac{2}{1+a^2} > 2-\dfrac{2}{a+b^2}\implies f(a) > f(b)\implies f$ is strictly increasing. Thus if we have $x > y > z \implies f(x) > f(y) > f(z)\implies y > z > x $, contradication to $x > z$. Thus at least two variables must be equal to each other, say $x = y \implies f(x) = f(y) \implies y = z \implies x = y =z\implies x = \dfrac{2x^2}{1+x^2}\implies x^3-2x^2+x = 0 \implies x(x^2-2x+1) = 0\implies x(x-1)^2 = 0 \implies x = 0, 1 \implies x = y = z =1$ or $x = y = z = 0$.

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