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. 

multiplying with $$a^3b^3c^3$$ then our inequality is equivalent to
$$a^8+b^8+c^8\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$ now we use the wellknown inequality
$$x^2+y^2+z^2\geq xy+yz+zx$$ thus we have
$$(a^4)^2+(b^4)^2+(c^4)^2\geq (ab)^4+(bc)^4+(ca)^4$$ and now again
$$((ab)^2)^2+((bc)^2)^2+((ca)^2)^2\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$
