For positive $a,b,c$ with $a\geq b \geq c$, show $$\sum_{\mbox{cyc}}\frac{3a^2+ac-c^2}{a^3+c^3}(a-c)^2 \geq \sum_{\mbox{cyc}}\frac{3a^2+ab-b^2}{a^3+b^3}(a-b)^2 \geq 0$$
Note that the two cyclic sums are not equivalent (even though they look to be identical with $b \leftrightarrow c$) because we have the convention that $$ \sum_{\mbox{cyc}} f(a,b,c) \equiv f(a,b,c) + f(b,c,a) + f(c,a,b)$$ which is not, in this case, the same as $f(a,c,b) + fc,b,a) + f(b,a,c)$.
Obviously, this is equivalent to saying that if $c\geq b \geq a >0$, then
$$\sum_{\mbox{cyc}}\frac{3a^2+ab-b^2}{a^3+b^3}(a-b)^2 \geq \sum_{\mbox{cyc}}\frac{3a^2+ac-c^2}{a^3+c^3}(a-c)^2 \geq 0$$
The $\sum_{\mbox{cyc}}\frac{3a^2+ac-c^2}{a^3+c^3}(a-c)^2 \geq 0$ part of this problem cropped up in the context of proving a slightly tougher inequality, and an unclear or invalid proof is offered on https://artofproblemsolving.com/community/c6h22937p427220 (at least, I don't see how the proof is valid).
I can prove either of the $\geq 0$ statements by letting $a = b+x = c+x+y$ with non-negative $x$ and $y$, writing out the sum, and clearing the denominators: The resulting degree $10$ multinomial ends up with no negative coefficients. I'm looking for a cleaner proof, and a proof that the ordering of the two sums is tied to whether the ordering of $a,b,c$ is an odd or even permutation of $a\geq b\geq c$
We need to prove that $$\sum_{cyc}\frac{(3b^2+ab-a^2)(a-b)^2}{a^3+b^3}\geq\sum_{cyc}\frac{(3a^2+ab-b^2)(a-b)^2}{a^3+b^3}$$ or $$\sum_{cyc}\frac{(4b^2-4a^2)(a-b)^2}{a^3+b^3}\geq0$$ or $$\sum_{cyc}\frac{(a-b)^3}{a^2-ab+b^2}\leq0$$ or $$\sum_{cyc}\frac{(a-b)(a^2-ab+b^2-ab)}{a^2-ab+b^2}\leq0$$ or $$\sum_{cyc}\frac{ab(a-b)}{a^2-ab+b^2}\geq0$$ $$\sum_{cyc}(a^4b^3-a^4c^3-a^4b^2c+a^4c^2b)\geq0$$ or $$(a-b)(a-c)(b-c)(a^2b^2+a^2c^2+b^2c^2)\geq0,$$ which is true for $a\geq b\geq c$.