<s>[INST] 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?
 [/INST] We need to prove that 
$$\sum_{cyc}(a^8-a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(2a^8-2a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(a^8-a^6b^2-a^2b^6+b^6+a^6c^2-2a^3b^3c^2+b^6c^2)\geq0$$ or
$$\sum_{cyc}((a^2-b^2)^2(a^4+a^2b^2+b^4)+c^2(a^3-b^3)^2)\geq0,$$
which is obvious.
More way.
By AM-GM:
$$\sum_{cyc}a^8=\frac{1}{8}\sum_{cyc}(3a^8+3b^8+2c^8)\geq$$
$$\geq\frac{1}{8}\sum_{cyc}8\sqrt[8]{a^{24}b^{24}c^{16}}=\sum_{cyc}a^3b^3c^2.$$
We can prove it by Rearrangement, which is the Chebyshov's result.
Indeed, by Rearrangement:
$$\frac{a^8+b^8+c^8}{a^3b^3c^3}=\frac{1}{2}\sum_{cyc}\left(2\cdot\frac{a^5}{b^3c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^5}{a^3b^3}+\frac{a^5}{a^3c^3}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{b^3}+\frac{a^2}{c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^3}+\frac{a^2}{a^3}\right)=\sum_{cyc}\frac{1}{a}.$$
</s>