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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:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$

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Just do the computation that Maple did by hand. Just mulpiply out. –  Rasmus Nov 27 '10 at 12:46
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Even if you expand out by hand, you won't recognize that the numerator is the square of this complicated expression. –  TCL Nov 27 '10 at 12:50
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If Maple did, you can too. Everything that Maple can do, a human can do. The thing is, Maple does it faster, that's all. –  Raskolnikov Nov 27 '10 at 13:25
    
Perhaps <a href="en.wikipedia.org/wiki/…; Muirhead's inequality</a> can help. –  TCL Nov 27 '10 at 14:00
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@Raskolnikov: "Everything that Maple can do, a human can do" may be true (even if the shortness of human lifespan makes it not very useful), but note that Maple isn't programmed to do only things obvious to all humans. To recognise that the numerator is a square, Maple may be using some non-trivial algorithms. (The algorithms by Gosper, Wilf-Zeilberger etc. as in the book "A=B" come to mind.) So "do it by hand" is not a complete answer to this question; it still needs to be explained to a human how to recognise that the numerator is a square. –  ShreevatsaR Jun 23 '11 at 4:55
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This problem may be amenable to the "Purkiss Principle". I'll leave it as an exercise for you to determine if it can be applied here. But even if not you should read the following beautiful article on it by Wm. Waterhouse Do Symmetric Problems Have Symmetric Solutions? I recall thinking that this was one of the most beautiful Monthly articles that I ever read as an undergraduate. Apparently others felt similarly since it won a prestigious Lester R. Ford award for expository excellence.

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Nice reference. Thanks! –  Andres Caicedo Nov 27 '10 at 21:38
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Thanks for the very nice reference! But of course it seems that the Purkiss symmetry principle is not very useful here: the expresssion is symmetric in $x, y, z$, but putting them equal gives zeroes in the denominator. (One could also say it's symmetric in $x-y$, $y-z$ and $z-x$, but that's saying the same thing.) –  ShreevatsaR Jun 23 '11 at 5:09
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It is not a solution, but a simplification. Claim: it is enough to consider the case $c,z=1$. Proof(sketch only because long): Denote $f(a,b,c,x,y,z):=\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2 - (c-a)^2-(c-b)^2$. Then $f(a,b,c,x,y,z)=f(a,b,c,\frac{x}{z},\frac{y}{z},1)$. So we have to investigate two cases, case 1) $z=1$, case 2) $z=0$.

Case 1) $z=1$. Define the function $f1(a,b,c,x,y):=\left(\frac{ax+by+c}{x-y}\right)^2+\left(\frac{ay+b+cx}{y-1}\right)^2+\left(\frac{a+bx+cy}{1-x}\right)^2 - (c-a)^2-(c-b)^2$. Then $c^2 f1(\frac{a}{c},\frac{b}{c},1,x,y)=f1(a,b,c,x,y)$ so we have two cases (i) $c=0$, (ii) $c=1$.

(i) $c=0$. Now we have two subcases, (a) $a=0$, (b) $a=1$.They are simply enough to handle by a human.

(ii) $c=1$. This is not so simple, but a human can handle this case also.

Case 2) $z=0$.We have two subcases, (i) $c=1$, (ii) $c=0$. The last one is again handable by a human.

So we have the case $c,z=1$. It is more complicated, but more or less we can say a human is able to recognize the complete square expression.

Edit: a confusing misprint is corrected. (Originally there was Case 1) $z=0$ Define the $\ldots$.)

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I don't understand why in Case 1) $z=0$, your $f1(a,b,c,x,y)\neq f(a,b,c,x,y,0)$? –  TCL May 24 '13 at 0:57
    
@TCL This corrected version can answer your question? –  vesszabo May 24 '13 at 10:10
    
I made one more correction. –  TCL May 24 '13 at 10:53
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