<s>[INST] Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$. 
Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.

Obviously, AM-GM seems to be suitable for LHS.
For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to do.
Can someone please teach me? Thank you.
p.s. That $\sqrt [27]{}$ is really terrible...
 [/INST] I found a solution that uses a bit of calculus.
First using the identity you have already found, we figure the inequality is equivalent to
$$\sqrt[3]{\frac{a^3+b^3+c^3+24}{27}} = \frac{a+b+c}{3} \le \sqrt[27]{\frac{a^3+b^3+c^3}{27}}.$$
Setting $x = \frac{a^3+b^3+c^3}{27}$ we find it equivalent to saying
$$(x+\frac89)^9-x \ge 0.$$
This is where the calculus enters the equation. It is a standard method to prove that all the zeros of $p(x) = (x+\frac89)^9-x$ are in fact negative. Since $x\ge0$ by definition this proves the inequality.
[The calculus: Take the derivative $p'(x) = 9(x+\frac89)^8-1$ which has two zeros $x_1,x_2$ smaller than zero. Show that $p(x_1),p(x_2)>0$.]
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