# Prove for $x$, $y$ and $z$

Well, last week I read a question and I really do not know what to do. I have tried everything I could think, but it does not seem to help me. So, if you can help me, I would be grateful. I'm not expecting answers, just hints - I don't want to see answers by now.

That's the question (it is from International Mathematical Olympiad. I read it translated to Portuguese, but I couldn't find the original one. So, I will try to translate to English again):

Let three positive numbers $x$, $y$ and $z$ $∈$ $\mathbb{R}$, and $xyz ≥ 1$. Prove that:

$$\frac{x^5 - x^2}{x^5 + y^2 + z^2} + \frac{y^5 - y^2}{y^5 + x^2 + z^2} + \frac{z^5 - z^2}{z^5 + x^2 + y^2} \geq 0.$$

Quickly, I thought:

$x^5 - x^2 = x^2(x - 1)(x^2 + x + 1)$;

$y^5 - y^2 = y^2(y - 1)(y^2 + y + 1)$;

$z^5 - z^2 = z^2(z - 1)(z^2 + z + 1)$.

But it doesn't seem to be really interesting to me. I think that might be more than this - a genious vision. And also, finding the LCM appears to be bruteforcing. No problems with that, but I think it can be simpler than that.

(I tagged it as a homework, but it's not exactly a homework. I find this question on my own.)

After all that stuff, I just have to thank you all. Sorry if I commited (m)any English mistakes, I am really tired and I wrote it as faster as I could to get your answers.

Thank you!

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If you combine the three fractions you get that you are trying to prove $\frac{x^5+y^5+z^5}{x^2+y^2+z^2}\ge 3$ – Ross Millikan Nov 15 '11 at 0:30
Really? I did not understand. Can you explain it better, please? – Ian Mateus Nov 15 '11 at 0:35
Related but different math.stackexchange.com/questions/61289/… – Henry Nov 15 '11 at 1:04
@Ross: Your expression does not seem to work with $x=y=z=1$ – Henry Nov 15 '11 at 1:48
@Henry: you are right. RHS should be $1$, not $3$ – Ross Millikan Nov 15 '11 at 2:53

Let, $A=x^{5}+y^{2}+z^{2}$

$B=y^{5}+x^{2}+z^{2}$

$C=z^{5}+x^{2}+y^{2}$

$\frac{[(y^{5}+x^{2}+z^{2})(z^{5}+x^{2}+y^{2})(x^{5}-x^{2})]+[(x^{5}+y^{2}+z^{2})(z^{5}+x^{2}+y^{2})(y^{5}-y^{2})]+[(x^{5}+y^{2}+z^{2})(y^{5}+x^{2}+z^{2})(z^{5}-z^{2})]}{ABC}\geq0$

$\frac{BC(x^{5}-x^{2})+AC(y^{5}-y^{2})+AB(z^{5}-z^{2})}{ABC}\geq0$

$\frac{[(y^{5}+x^{2}+z^{2})(z^{5}+x^{2}+y^{2})(x^{2})(x^{3}-1)]+[(x^{5}+y^{2}+z^{2})(z^{5}+x^{2}+y^{2})(y^{2})(y^{3}-1))]+[(x^{5}+y^{2}+z^{2})(y^{5}+x^{2}+z^{2})(z^{2})(z^{3}-1)]}{(x^{5}+y^{2}+z^{2})(y^{5}+x^{2}+z^{2})(z^{5}+x^{2}+y^{2})}\geq0$

This has become a tedious computation. So going back to the A,B,C one can attempt grouping in order to simplify:

$\frac{BCx^{5}-BCx^{2}+ACy^{5}-ACy^{2}+ABz^{5}-ABz^{2}}{ABC} \geq0$

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