Problem (KaiRain's problem).
For $a,b,c\geqslant 0.$ Prove $$\displaystyle \sum_{\text{cyc}}\, (23a-5b-c)(a-b)^2(a+b-3c)^2 \geqslant 0$$
I only found a proof by $pqr.$ (Note that from pqr's proof we can get SOS but very ugly and my SOS solution is from that so I will not post my SOS)
Need to prove $$18{p}^{5}-136{p}^{3}q+34{p}^{2}r+288p{q}^{2}-480qr \geqslant 14\left( b-c \right) \left( c-a \right) \left( a-b \right) ( 5{p}^{2}-16q) $$ After squaring both side$,$ it's $$ \left( 133456\,{p}^{4}-879360\,{p}^{2}q+1585152\,{q}^{2} \right) {r}^ {2}+$$ $$+8\,p \left( 2603\,{p}^{6}-30021\,{p}^{4}q+114416\,{p}^{2}{q}^{2}- 147456\,{q}^{3} \right) r+$$ $$+4\, \left( 9\,{p}^{6}-40\,{p}^{4}q-17\,{p}^{ 2}{q}^{2}+196\,{q}^{3} \right) \left(3{p}^{2} -16q \right) ^{2} \geqslant 0$$
Since $8341p^4-54960p^2q+99072q^2 \geqslant 0.$
If $133p^2-688q >0$ then $$2603\,{p}^{6}-30021\,{p}^{4}q+114416\,{p}^{2}{q}^{2}-147456\,{q}^{3} \geqslant 0, 9\,{p}^{6}-40\,{p}^{4}q-17\,{p}^{2}{q}^{2}+196\,{q}^{3} \geqslant 0$$ then qed.
Else$,$ we have $$\Delta_r =3136\, \left( 133\,{p}^{2}-688\,q \right) \left( {p}^{2}-3\,q \right) ^{2} \left( 5\,{p}^{2}-16\,q \right) ^{4}\leqslant 0.$$
Done.
But any other proof$?$