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 :-)

General comment: as soon as one tries to use AM-GM for 
$x=a^3+a+1,$ $y=b^3+b+1$ and $z=c^3+c+1$ the inequality becomes wrong, since
$$a^3+a+b^3+b+c^3+c+3\ge 2(a^2+b^2+c^2)+3=9.$$
Using Lagrange multiplayers, one can reduce this problem to the following system:
$$(3a^2+1)(b^3+b+1)(c^3+c+1)=2\lambda a$$ 
$$(3b^2+1)(a^3+a+1)(c^3+c+1)=2\lambda b$$ 
$$(3c^2+1)(a^3+a+1)(b^3+b+1)=2\lambda c.$$
In other words, if $\lambda\ne 0,$ for the function
$$f(x)=\frac{x(x^3+x+1)}{(3x^2+1)}$$
 we have $f(a)=f(b)=f(c).$ It is easy to see, that $f$ is monotone for $x\ge 0$ so the only option is $a=b=c=1.$ The rest should be clear.
