Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$ 
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
  Prove that: 
  $$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$

I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and 
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing. 
Thanks :-)

Let $u:=a^2, v:=b^2, w:=c^2$, we have $u+v+w=3$.
Consider the function
$$f(x)=\ln (1+x^{\frac{1}{2}}+x^{\frac{3}{2}}),\ 0<x\leq 3$$
it's easy to compute that $f''(x)<0$.
by Jensen's inequality, we have
$$\sum\ln (1+u^{\frac{1}{2}}+u^{\frac{3}{2}})\leq3f(\dfrac{\sum u}{3})=3\ln 3$$
that is
$$\prod(a^3+a+1) \leq 27$$
When the max occurs, we have $u=v=w\Rightarrow a=b=c$
Q.E.D.
By the way, I draw a graph of $f''(x)$ on $(0,3]$ by mathematica to show it more directly...

