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.

Geometric approach :


Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :
    $$T\leq\frac{3\sqrt{3}abc}{4(a+b+c)}$$ 
    Or $$4\sqrt{3}T\leq \frac{9abc}{a+b+c}$$
    Where $T$ is the area of the triangle $ABC$


For a proof see (found on Wikipedia):
Posamentier, Alfred S. and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.
