Two inequalities involving the rearrangement inequality

by AM-GM we get $$\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2}{6}\geq \sqrt[6]{a^{12}b^{12}c^{12}}=a^2b^2c^2$$

Without loss of generality, $a\le b\le c$. Then $ab\le ac\le bc$. Denote $x_1=x_2=ab, x_3=x_4=ac, x_5=x_6=bc$, and $y_1=y_2=a,y_3=y_4=b,y_5=y_6=c$. Then both $x_i$ and $y_i$ are monotonic, and $x_1y_6+...+x_6y_1=6abc$, so it is the least sum among all rearrangements. It is not hard to find a rearrangement to get left-hand side of the second inequality. The first is the same, just change all the letters by their squares.

$(a,b,c)$ 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!