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 $$((ab)^2)^2+((bc)^2)^2+((ca)^2)^2\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$

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 $$ a^8+b^8+c^8\ge ab+bc+ca.$$

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.