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?
 A: Use the inequality $x^2+y^2+z^2>zy+yz+zx$ for distinct $x,y,z$ repeatedly: $$a^8+b^8+c^8>a^4b^4+b^4c^4+c^4a^4>a^4b^2c^2+a^2b^4c^2+a^2b^2c^4>a^3b^3c^2+a^2b^3c^3+a^3b^2c^3.$$
A: 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 $a=b=c$.
A: 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.
A: This is the same as
$$a^8+b^8+c^8> a^2b^3c^3+a^3b^2c^3+a^3b^3c^2.$$
You can get this by using AM/GM to get an upper bound for each term
on the right.
A: 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:
$$\frac{a^8+b^8+c^8}{a^3b^3c^3}=\frac{1}{2}\sum_{cyc}\left(2\cdot\frac{a^5}{b^3c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^5}{a^3b^3}+\frac{a^5}{a^3c^3}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{b^3}+\frac{a^2}{c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^3}+\frac{a^2}{a^3}\right)=\sum_{cyc}\frac{1}{a}.$$
A: 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.  
