<s>[INST] 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 :-)
 [/INST] Final solution : 
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=527572&p=2997801#p2997801
$$ (a-1)^4\geq0\Rightarrow a^3+a+1\leq\frac{(a^2+1)(a^2+5)}{4} $$
$$ \prod(a^3+a+1)\leq\frac{1}{64}\prod(a^2+1)\prod(a^2+5)$$
$$ \leq\frac{1}{64}\frac{(a^2+b^2+c^2+3)^3}{27}\frac{(a^2+b^2+c^2+15)^3}{27}=27 $$
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