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. 

Multiply everywhere by $a^3b^3c^3$ to get that we want to prove
$$
a^8 + b^8 + c^8 \geq a^2b^3c^3 + a^3b^2c^3 + a^3b^3c^2
$$
Let's start with the right-hand side, and assume that $a\leq b\leq c$. In that case, $a^3b^3\leq a^3c^3\leq b^3c^3$ and $a^2\leq b^2\leq c^2$, so by the rearrangement inequality, we have
$$
a^2b^3c^3 + a^3b^2c^3 + a^3b^3c^2 \leq a^2a^3b^3 + b^2a^3c^3 + c^2c^3b^3\\
= a^5b^3 + a^3b^2c^3 + c^5b^3
$$
And again, by the rearrangement inequality (twice, in brackets), we have
$$\bigg[a^5b^3 + a^3b^2c^3\bigg] + c^5b^3\\
\leq \bigg[a^5a^3 + b^3b^2c^3\bigg] + c^5b^3 \\
= a^8 + \bigg[b^5c^3+c^5b^3\bigg]\\
\leq a^8 +  \bigg[b^5b^3 + c^5c^3\bigg] \\
=a^8 + b^8+c^8$$ This finishes our proof.
