<s>[INST] 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:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$
 [/INST] 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.
</s>