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?

On the LHS you have $$\frac{a^5}{b^3c^3} + \frac{b^5}{a^3c^3} +
 \frac{c^5}{a^3b^3}$$
Relabel this as $$x_1 + x_2 + x_3$$
The standard way to proceed with AM-GM is to fiddle around with products of $x_1, x_2, x_3$ until you find some $\sqrt[n_1 + n_2 + n_3]{x_1^{n_1}x_2^{n_2}x_3^{n_3}}$ that yields one of the RHS terms.  
In our case a little fiddling gives $$x_1^3x_2^3x_3^2 = \frac{1}{c^8}$$
So we have by AM-GM 
$$\frac{3x_1 + 3x_2 + 2x_3}{8} \geq \sqrt[8]{x_1^3x_2^3x_3^2} = \frac{1}{c}$$
And similarly 
$$\frac{3x_1 + 2x_2 + 3x_3}{8} \geq \frac{1}{b}$$
$$\frac{2x_1 + 3x_2 + 3x_3}{8} \geq \frac{1}{a}$$
Adding these three inequalities together gives you the one you set out to prove.  
