Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$ For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4\,{p}^{4}q+8\,{p}^{3}r+27\,{r}^{2} \geqq 0$$
Or$:$ $${\frac { \left( {p}^{4}-5\,{p}^{2}q+6\,pr+4\,{q}^{2} \right) 
 \left( 7\,{p}^{4}+45\,{p}^{2}q+54\,pr-36\,{q}^{2} \right) }{12{p}^{2}}}
+\,{\frac { \left( {p}^{2}-3\,q \right)  \left( 5\,{p}^{2}-3\,q
 \right)  \left( {p}^{2}-4\,q \right) ^{2}}{12{p}^{2}}} \geqq 0$$
Which is obvious because $p^2 \geqq 3q,\, p^4 -5p^2 q+6pr+4q^2 \geqq 0 \,(\text{Schur degree 4})$
I hope for another proof (without $uvw$!). Thanks for a real lot!
PS$:$ You can get $pqr$'s form more faster by using Maple$,$ see here.

If $a+b-c<0$ and $a+c-b<0$, we obtain $a<0,$ which is a contradiction.
Thus, it's enough to prove our inequality for $a+b-c>0$, $a+c-b>0$ and $b+c-a>0.$
Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2\geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz).$$
Can you end it now?
