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 inequality is equivalent to 
$$a^8 + b^8 + c^8 > a^2b^3c^3 + a^3b^2c^3+ a^3b^3c^2.$$
Now apply Muirhead's inequality (i.e. $[8,0,0]\geq [3,3,2]$) and note that the equality holds iff $a=b=c$.
