Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$ As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$. 
Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead's inequality, yet I ought to prove it using the rearrangement inequality. I can't however figure out any suitable sequences and permutations of them. 

(Added a solution as it's much simpler than the rest.)
Multiplying throughout by $a^3b^3c^3$, we WTS
$$a^8+b^8+c^8\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2.$$
This follows directly by summing up the cyclic inequalities of
$$ 2a^8 + 3b^8 + 3c^8 \geq 8 a^2b^3c^3.$$
