<s>[INST] 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. 
 [/INST] $(a^4,b^4,c^4)$ and $(a^4,b^4,c^4)$ are the same ordered.
Thus, $a^8+b^8+c^8=a^4\cdot a^4+b^4\cdot b^4+c^4\cdot c^4\geq a^4b^4+b^4c^4+c^4a^4$ and it remains to prove that
$$\sum_{cyc}a^4b^4\geq \sum_{cyc}a^2b^3c^3$$ or
$$\sum_{cyc}\frac{a^2b^2}{c^2}\geq\sum_{cyc}ab.$$
Since $(a^2b^2,a^2c^2,b^2a^2)$ and $\left(\frac{1}{c^2},\frac{1}{b^2},\frac{1}{a^2}\right)$ are the same ordered, by Rearrangement we obtain:
$$\sum_{cyc}\frac{a^2b^2}{c^2}\geq\sum_{cyc}\frac{a^2b^2}{b^2}=\sum_{cyc}a^2\geq\sum_{cyc}ab,$$
where the last inequality it's Rearrangement again.
Done! 
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