<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] Without loss of generality, let $abc=1$. Then the inequality becomes
$$ a^8+b^8+c^8\ge ab+bc+ca.$$
Note
\begin{eqnarray}
2(a^8+b^8+c^8)&=&(a^8+b^8)+(b^8+c^8)+(c^8+a^8)\\
&\ge&2a^4b^4+2b^4c^4+2c^4a^4\\
&=&(a^4b^4+b^4c^4)+(b^4c^4+c^4a^4)+(c^4a^4+a^4b^4)\\
&\ge&2a^2b^4c^2+2a^2b^2c^4+2a^4b^2c^2\\
&=&2a^2+2b^2+2c^2\\
&=&(a^2+b^2)+(b^2+c^2)+(c^2+a^2)\\
&\ge&2ab+2bc+2ca
\end{eqnarray}
and hence
$$ a^8+b^8+c^8\ge ab+bc+ca.$$
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