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Let $a,b,c,x,y,z\in\mathbb{R}$. Prove that $$ \left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\geq(c-a)^2+(c-b)^2$$

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    @Jaska. Damn; I was sure I had already passed this course. Why am I still getting assignments?2010-11-23
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    Because I don't know how to solve it.2010-11-23
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    Well, simplify the LHS and then the answer will stare at you for a while.2010-11-23
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    @Jaska: So, since *you* can't solve it, you are ordering me to solve it? How about *asking* for help, and saying what you've tried, instead of using the Imperative Mode and telling us what to do?2010-11-23
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    OK. Well first I assumed that $xyz\ne 0$. Then I tried to show that LHS-RHS=(a x^2 y-3 a x y z+a x z^2+a y^2 z+b x^2 z+b x y^2-3 b x y z+b y z^2+c x^3-c x^2 y-c x^2 z-c x y^2+3 c x y z-c x z^2+c y^3-c y^2 z-c y z^2+c z^3)^2/((x-y)^2 (x-z)^2 (z-y)^2) but I haven't found a clever way to show it.2010-11-23
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    @Jaska: Why don't you edit the question with what your thoughts were? Also, what is source of this? This seems pointless to me. What is the motivation for this inequality?2010-11-23
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    A friend of mine asked hints how to learn to solve maths olympics problems. I found no answer how to get intuition to solve inequalities like this.2010-11-23
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    @Jaska: Edit the question and add the information to the body, and try to avoid coming off as if you were assigning problems to the readers of this site.2010-11-24
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    @Jaska: Your profile says "zero votes". Now I have even less motivation to help you train for the Olympiad.2010-11-24
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    I have misunderstood the idea of the forum. A friend of mine asked problem, I couldn't solve it so I was looking someone who could done it. What kind of questions I should ask here and how to ask them? And what are votes? As I see my profile, there are many questions with votes.2010-11-24
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    @Yuval: Unregistered users cannot vote. Doesn't it help that Jaska accepts answers? @Jaska: The votes come from (registered) users deciding to vote up or down on questions and answers, which helps to encourage high quality contributions to the site. It is not something to be taken too seriously, but it can also be a good thing to vote up answers to your questions if they are helpful.2010-11-26
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    @Jaska: As for the kind of questions and how to ask them, this is an appropriate question. However, keep in mind that you are asking people to take time to help you, and most of us appreciate pleasant wording and indications of effort that went into the problem before posting. It would also be better to include the source of the problem, as Moron indicated.2010-11-26
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    I found this unsolved in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=293472&p=1586692#p1586692 .2010-11-27

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With help from Maple, I got $$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$ equal to $$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$.

But with no help from a computer algebra, how would one prove this?