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

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 $\ldots$.)
