Find max: $\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$ For $a,b,c>0$ and $abc=1$. Find max:
$\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$
I used AM-GM for $a^3+2b^3$ but I don't know how to continue ...

From the AM-GM Inequality, 
$a^3+b^3+1\ge 3ab$,
$b^3+1+1\ge 3b$
$\Rightarrow a^3+2b^3+6\ge 3(ab+b+1)$
Similarly, $b^3+2c^3+6\ge 3(bc+c+1)$,$c^3+2a^3+6\ge 3(ca+a+1)$.
We have $\frac 1{a^3+2b^3+6}+\frac 1{b^3+2c^3+6}+$$\frac 1{c^3+2a^3+6}\le \frac 13(\frac 1{ab+b+1}+\frac 1{bc+c+1}+\frac 1{ca+a+1})$
$=\frac 13(\frac 1{ab+b+1}+\frac {ab}{b+1+ab}+\frac{b}{1+ab+b})$
$=\frac 13$.
The equality occurs when $a=b=c=1$
