Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$

Question

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.

Answer 1

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?

Answer 2

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.