<s>[INST] 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?
 [/INST] 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,
$$2(a^3+b^3+c^3> a^2(b+c)+b^2(c+a)+c^2(a+b)\geq 6abc $$
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