If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?

The given inequality can be written as
$$a^8+b^8+c^8 > a^2b^3c^3+b^2a^3c^3+c^2a^3b^3.$$
Since $(8,0,0)$ majorizes $(2,3,3)$, therefore we can apply Muirhead's inequality to get the result.
