How to prove this inequality without use of computers? 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:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$
 A: 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.
A: We may express it in matrix form. Let $u = [a, \ b, \ c]^\mathsf{T}$,
$p_1 = [x, \ y, \ z]^\mathsf{T}$, $p_2 = [y, \ z, \ x]^\mathsf{T}$,
$p_3 = [z, \ x, \ y]^\mathsf{T}$,
$q_1 = [-1, \ 0, \ 1]^\mathsf{T}$ and $q_2 = [0, \ -1, \ 1]^\mathsf{T}$.
Let
$$S = \frac{1}{(x-y)^2} p_1p_1^\mathsf{T}
+ \frac{1}{(y-z)^2} p_2p_2^\mathsf{T} + \frac{1}{(z-x)^2} p_3p_3^\mathsf{T}
- q_1q_1^\mathsf{T} - q_2q_2^\mathsf{T}.$$
We have $\mathrm{LHS} - \mathrm{RHS} = u^\mathsf{T}S u$.
It suffices to prove that $S$ is positive semidefinite.
Note that all $2\times 2$ minors of $S$ are zero
(e.g., $S_{1,1} S_{2,2}-S_{1,2}S_{2,1} = 0$, etc.). Thus, $\mathrm{rank}(S)\le 1$.
Also, $S_{1,1} = x^2/(x-y)^2+y^2/(y-z)^2+z^2/(z-x)^2-1 > 0$
(the proof is not difficult).
Thus, $S$ is positive semidefinite. We are done.
A: 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 $\ldots$.)
A: This solution should be considered reasonable. First observe that both sides of the inequality is quadratic in $c,$ and expanding to get explicit forms of those quadratics is not difficult. Namely, the RHS is:
$$R(c) = 2c^2 - 2c(a+b) + a^2+b^2.$$
The LHS is a bit more tedious so let's compute the coefficients separately. Write $LHS:=L(c) = Ac^2+2Bc+Cc$ and then:
$$\begin{cases}
A = \dfrac{x^2}{(y-z)^2} + \dfrac{y^2}{(z-x)^2}+\dfrac{z^2}{(x-y)^2}\\
2B = 2\sum\limits_{cyc}\dfrac{z}{(x-y)^2}(ax+by)\\
C = \sum{cyc}\dfrac{(ax+by)^2}{(x-y)^2}
\end{cases}.$$
First, we tackle $A.$ Actually, it is sort of a well-known identity/inequality as:
$$A - 2=\left(\dfrac{x}{y-z}+\dfrac{y}{z-x}+\dfrac{z}{x-y}\right)^2\geq 0,\quad (1).$$
I leave the solution to the above as a simple exercise. The more interesting one is $C,$ which basically was solved as a separate problem in this thread:
$$\sum_{x,y,z}\left(\dfrac{mx-ny}{x-y}\right)^2=\Big(\dfrac{my-nz}{y-z} - \dfrac{x(m-n)(z-y)}{(x-y)(z-x)}\Big)^2+m^2+n^2.$$ Taking $m = a$ and $n = -b$ will give us:
$$C = a^2+b^2 + \Delta ^2,\quad (2).$$
where $\Delta$ can be calculated from the linked answer.
Now let us recollect what we are trying to prove. we are trying to prove that the following is a non-negative quadratic:
$$L(c) - R(c) = (A - 2)c^2 + 2(B +a+b )c + (C-a^2 - b^2)\geq 0.$$
But this is equivalent to having a zero-discriminant and a non-negative leading coefficient, the latter of which is already checked $(1).$ From $(1),$ let's let $A - 2 = \Omega^2,$ then the discriminant condition is:
$$4(B-a-b)^2 = 4(A-2)(C-a^2-b^2)\iff B+a+b = \pm\Omega\Delta,\quad  (3).$$
But now this can be checked by hand since:
$$\begin{cases}
B+a+b = a\left(\dfrac{zx}{(x-y)^2}-1\right)+b\left(\dfrac{zy}{(x-y)^2}-1\right)\\
\Omega = \sum\dfrac{x}{y-z} \\
\Delta = a\left(\dfrac{y}{y-z} - \dfrac{x}{x-y}-\dfrac{x}{z-x}\right) + b\left(\dfrac{z}{y-z} - \dfrac{x}{x-y}-\dfrac{x}{z-x}\right)
\end{cases}
$$
