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. 
 A: 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
$$((ab)^2)^2+((bc)^2)^2+((ca)^2)^2\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$
A: 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
$$ a^8+b^8+c^8\ge ab+bc+ca.$$
A: 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.
A: $(a^4,b^4,c^4)$ 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! 
A: (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
$$ 2a^8 + 3b^8 + 3c^8 \geq 8 a^2b^3c^3.$$
