Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$ 
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and 
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope there'a an exploitable link between them. They fall easily under Muirhead, yet I cannot figure out how to prove them using the Rearrangement Inequality. 
Any hints greatly appreciated. 

$(a,b,c)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered.
Thus, by Rearrangement $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a\cdot\frac{1}{a}+b\cdot\frac{1}{b}+c\cdot\frac{1}{c}=3,$$ which gives 
$$a^2c+b^2a+c^2b\geq3abc$$
Similarly we'll get $$a^2b+b^2c+c^2a\geq3abc$$ and after summing we are done!
