<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] Let $s = a+b+c$. Then, as you noted, the inequality is equivalent to:
$$ \frac{s}{3} \geq \sqrt[27]{\frac{s^3-24}{3}} \qquad (\ast)$$
Because you know that $(a+b)(b+c)(c+a) = 8$, or equivalently $\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2} = 1$, it follows that $s = \frac{a+b}{2} + \frac{b+c}{2} + \frac{c+a}{2} \geq 3$. In fact, this is all you can say about $s$, so our initial problem is equivalent to showing that $(\ast)$ holds for all $s \geq 3$. You can rewrite $(\ast)$ as:
$$ \left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3} \geq 0 \qquad (\ast\ast)$$
 For $s = 3$ there is equality in $(\ast\ast)$, and we can in fact divide out $(s-3)^2$ from the left side. The resulting function:
$$ \frac{\left( \frac{s}{3} \right)^2 - \frac{s^3-24}{3}}{(s-3)^2} $$
is a polynomial which takes strictly positive values for $s \geq 3$ (in fact, for $s \geq -3$). You can directly verify that by a variety of brute-force methods (possibly with help of some softwere like Mathematica)

A more satisfactory idea: Put $t = (s/3)^3 \geq 1$ for ease of notation. We can easily see that $(\ast\ast)$ is equivalent to $$t^{9} - 9t + 8 \geq 0 $$
But this follows directly by applying AM-GM to $t^9,1,1,1,1,1,1,1,1$!. 
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