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
 A: Let $f(x,y,z) = (x^3 +1)(y^3 +1)(z^3 +1)$ subject to the constraint $g(x,y,z) = x + y + z - 3 =0$ and $ x,y,z \ge 0$. 
Then using lagrange multipliers we have


*

*$f_x = 3x^2(y^3 +1)(z^3 +1)= g_x= \lambda$

*$f_y = 3y^2(x^3 +1)(z^3 +1)= g_y = \lambda$

*$f_z = 3z^2(x^3 +1)(y^3 +1)=g_z = \lambda$

*$g(x,y,z) = x + y + z - 3 =0$


From the first three equations we see $x^2(y^3 +1) = y^2(x^3 +1) = z^2(y^3 +1)$. So then $x =z$, since $x, z \ge 0$. Then clearly $x=y=z$, so plugging into our formula for $g$, we see $x=y=z=1$, so $(1,1,1)$ is a critical point.
Now, $f(1,1,1) = 8$. To see this is a minimum, we see $(0,0,3)$ lies on our curve, and $f(0,0,3) = 28 \gt f(1,1,1) = 8.$
We conclude that $(a^3 +1)(b^3 +1)(c^3 +1) \ge 8$.
A: $\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2\right)=\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2-\frac{3}{2}(a-1)\right)$.
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!
A: 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 $$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
