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

BY Jensen inequality with  $\phi\left(t\right)=\ln(t^3+1)$
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $a+b+c= 3$ we get 
$$\ln(2)=\phi\left(1\right)\le \frac13\left(\ln(a^3+1)+\ln(b^3+1)+\ln(c^3+1)\right) $$
hence $$\ln8= 3\ln(2)\le \ln \left((a^3+1)(b^3+1)(c^3+1)\right)$$
that is $$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
