prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so 
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think  this method is wrong

$\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2\right)=\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2-\frac{3}{2}(a-1)\right)$.
Let $f(x)=\ln(x^3+1)-\ln2-\frac{3}{2}(x-1)$.
Easy to see that $\min\limits_{[0,2]}f=0$.
Indeed, $f'(x)=\frac{3x^2}{x^3+1}-\frac{3}{2}=\frac{3(1-x)(x^2-x-1)}{2(x^3+1)}$ and since $x_{max}=\frac{1+\sqrt5}{2}$, 
it remains to check that $f(2)>0$, which is true.
Thus, for $\{a,b,c\}\subset[0,2]$ our inequality is proven.
Let $a>2$.
Hence, $\prod\limits_{cyc}(a^3+1)\geq(8+1)\cdot1\cdot1>8$.
Done!
