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

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:
$$\begin{cases}
B+a+b = a\left(\dfrac{zx}{(x-y)^2}-1\right)+b\left(\dfrac{zy}{(x-y)^2}-1\right)\\
\Omega = \sum\dfrac{x}{y-z} \\
\Delta = a\left(\dfrac{y}{y-z} - \dfrac{x}{x-y}-\dfrac{x}{z-x}\right) + b\left(\dfrac{z}{y-z} - \dfrac{x}{x-y}-\dfrac{x}{z-x}\right)
\end{cases}
$$
