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 $$27(x+y)^2(x+z)^2(y+z)^2geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz).$$

Can you end it now?

Geometric approach :

Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :

$$Tleqfrac{3sqrt{3}abc}{4(a+b+c)}$$

Or $$4sqrt{3}Tleq frac{9abc}{a+b+c}$$

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.

Suppose $a=max{a,b,c}.$

If $a geqslant b + c$ then

$$text{LHS} leqslant 0 leqslant text{RHS}.$$

If $a leqslant b + c,$ we write inequality as

$$ 27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) cdot (a+b+c)^3 leqslant 9^3 cdot a^3b^3c^3.$$

By the AM-GM inequality we have

$$27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) leqslant left[a(b+c-a)+b(c+a-b)+c(a+b-c)right]^3.$$

So, we need to prove

$$(a+b+c)left[2(ab+bc+ca)-a^2-b^2-c^2right] leqslant 9abc.$$

It’s Schur inequality.