Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that  

$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$  

How can I show this?
 A: *

*By my answer here and with the condition given,



$$2\left(a^3+b^3+c^3\right)>2\left(a^2b+b^2c+c^2a\right)$$

$\left(a^2b+b^2c+c^2a\right)>\left(ab^2+bc^2+ca^2\right)\\\impliedby ab(a-b)+bc(b-c)-ca(a-b)-ca(b-c)>0\\ \impliedby a(b-c)(a-b)-c(a-b)(b-c)>0 \\\impliedby (a-c)(a-b)(b-c)>0$
$$\color{blue}{\boxed{\color{blue}{\therefore2\left(a^3+b^3+c^3\right)>2\left(a^2b+b^2c+c^2a\right)>ab(a+b)+bc(b+c)+ca(c+a)}}}$$


*$ab^2+bc^2+ca^2>3abc\\\impliedby ab\left(b-c\right)+bc\left(c-a\right)+ca\left(a-b\right)>0\\\impliedby ab\left(b-c\right)-bc\left(a-b\right)-bc\left(b-c\right)+ca\left(a-b\right)>0\\\impliedby b(a-c)(b-c)+c(a-b)^2>0$


$$\color{blue}{\boxed{\color{blue}{\therefore ab(a+b)+bc(b+c)+ca(c+a)>6abc}}}$$
A: Using Rearrangement inequality for $a,a^2$ and $b,b^2$
$$a^3+b^3\gt a^2b+b^2a$$
Similarly,
$$b^3+c^3\gt b^2c+c^2b$$
and   
$$c^3+a^3\gt c^2a+a^2c$$
Adding these three, we have  
Using sing AM-GM, we have
$$2(a^3+b^3+c^3)> a^2b+a^2c+b^2c+b^2a+c^2a+c^2b$$
$$\dfrac{a^2b+a^2c+b^2c+b^2a+c^2a+c^2b}{6}\ge\sqrt[6]{a^6b^6c^6}$$
$$\implies a^2b+a^2c+b^2c+b^2a+c^2a+c^2b\geq 6abc $$
Finally,
$$2(a^3+b^3+c^3> a^2(b+c)+b^2(c+a)+c^2(a+b)\geq 6abc $$
A: Hint: $a^3 + b^3 > a^2b + ab^2$,
and use AM-GM inequality for $6$ terms for the other one.
