For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} +\frac{2(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \leqslant \frac{(a+b+c)^2}{ab+bc+ca} $$ I found it when I tried to find the stronger version of this, you can see my post in #4.
This is true, which is verify by Maple. So I tried to get in pqr form, but it's so hard. $$20736\,q{r}^{4}+96\,pq \left( 59\,{p}^{2}-306\,q \right) {r}^{3}+ \left( 8\,{p}^{8}+380\,{p}^{6}q-3984\,{p}^{4}{q}^{2}+9600\,{p}^{2}{q} ^{3}+5664\,{q}^{4} \right) {r}^{2}+p \left( {p}^{10}-6\,{p}^{8}q+11\,{ p}^{6}{q}^{2}-132\,{p}^{4}{q}^{3}+1344\,{p}^{2}{q}^{4}-4032\,{q}^{5} \right) r+6\,{q}^{5} \left( {p}^{2}-8\,q \right) ^{2} \geqslant 0,$$ Where $p=a+b+c,q=ab+bc+ca,r=abc.$
BW kills it immediately but it's not available with no computer.
So I hope to another solution. Of course, except BW, any solution are welcome here include SOS.
Thanks for a real lot!