# Second degree polynomial inequality

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? – Arturo Magidin Nov 23 '10 at 21:53
@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? – Arturo Magidin Nov 23 '10 at 21:57
@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? – Aryabhata Nov 23 '10 at 22:42
@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. – Arturo Magidin Nov 24 '10 at 2:33
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 $\ge 0$.