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 :-)
 A: For what it's worth, this answer uses derivatives, but it does use a generally applicable method.
Since $a^2+b^2+c^2=3$, any variations, $(\delta a,\delta b,\delta c)$, of $(a,b,c)$ must satisfy
$$
a\,\delta a+b\,\delta b+c\,\delta c=0\tag{1}
$$
We are interested in finding the maximum of
$$
\log(a^3+a+1)+\log(b^3+b+1)+\log(c^3+c+1)\tag{2}
$$
At a critical point, the variations of $(2)$ must satisfy
$$
\frac{3a^2+1}{a^3+a+1}\delta a+\frac{3b^2+1}{b^3+b+1}\delta b+\frac{3c^2+1}{c^3+c+1}\delta c=0\tag{3}
$$
Standard linearity arguments say that if $(3)$ is true for all $(\delta a,\delta b,\delta c)$ that satisfy $(1)$, we have
$$
\left(\frac{3a^2+1}{a^3+a+1},\frac{3b^2+1}{b^3+b+1},\frac{3c^2+1}{c^3+c+1}\right)=k(a,b,c)\tag{4}
$$
That is,
$$
\frac{3a^2+1}{a^4+a^2+a}=\frac{3b^2+1}{b^4+b^2+b}=\frac{3c^2+1}{c^4+c^2+c}\tag{5}
$$
Note that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\frac{3x^2+1}{x^4+x^2+x}
=-\frac{6 x^5+4 x^3-3 x^2+2 x+1}{(x^4+x^2+x)^2}\tag{6}
$$
If $x\ge1$, then $6x^5+4x^3\ge3x^2$ and if $0\le x\le1$, then $2x+1\ge3x^2$. Therefore, for all $x\ge0$, $(6)$ is negative. That is,
$$
\frac{3x^2+1}{x^4+x^2+x}\tag{7}
$$
is monotonic decreasing which, when combined with $(5)$, says that
$$
a=b=c\tag{8}
$$
$(8)$ says that
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=27\tag{9}
$$
Condition $(8)$ assumes that (a,b,c) is not on the boundary, that is none are $0$. Suppose that $c=0$, then the same argument yields that $a=b=\frac12\sqrt6$ and therefore
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=\frac{83}{8}+\frac52\sqrt6\tag{10}
$$
Suppose that $b=c=0$, then $a=\sqrt3$ and therefore
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=1+4\sqrt3\tag{11}
$$
Comparing $(9)$, $(10)$, and $(11)$, the maximum is $27$.
A: 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.
A: This will have derivatives. Substitute $a=\sqrt{x}, b=\sqrt{y}$ and $c=\sqrt{z}$. Then $x+y+z=3$. Consider the function $f(x)=\ln(x\sqrt{x}+\sqrt{x}+1)$. It is concave.
Hence Jensen yields: $f(x)+f(y)+f(z)<=3f(1)=3\ln{3}$
This is equivalent to what is asked.
A: I have a new answer. We will apply the Cebysev's inequality. 
$$(a^3+a+1)(b^3+b+1)\leq3(a^3b^3+ab+1)$$ 
So: 
$$\left[(a^3+a+1)(b^3+b+1)(c^3+c+1)\right]^{2} \leq 27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1).$$
Now, we will prove that: 
$$27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27^{2}.$$ Equivalent with:
$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27.$$
Now we use $AM \geq GM$ :
$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq \left(\frac{a^3b^3+b^3c^3+a^3c^3+ab+bc+ca+3}{3}\right)^{3}.$$
Now we will apply the following inequality to obtain the desired result: 
$$a^3b^3+b^3c^3+a^3c^3 \leq \frac{(a^2+b^2+c^2)^{3}}{9}=3,$$
$$ab+bc+ca \leq a^2+b^2+c^2=3.$$
A: It is enough to show that for $t\geqslant0$,
$$f(t) = \log(3)-\log\left(t^3+t+1\right)+\tfrac23\left(t^2-1\right)
\geqslant 0$$
as the inequality in question is merely $f(a)+f(b)+f(c)\geqslant0$.
But $f’(t)=0$ only when $t=1$ where it has a minimum, so $f(t)\geqslant f(1)=0$.
A: 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...

A: Final solution : 
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=527572&p=2997801#p2997801
$$ (a-1)^4\geq0\Rightarrow a^3+a+1\leq\frac{(a^2+1)(a^2+5)}{4} $$
$$ \prod(a^3+a+1)\leq\frac{1}{64}\prod(a^2+1)\prod(a^2+5)$$
$$ \leq\frac{1}{64}\frac{(a^2+b^2+c^2+3)^3}{27}\frac{(a^2+b^2+c^2+15)^3}{27}=27 $$
A: This is wrong
If you are familiar with majorization, observe that the function is schur concave. Thus, its maximum occurs at a point where all variables are equal, and since that point exists in the constraint set (i.e. $a,b,c\geq 0$ and $a^2+b^2+c^2=3$), $a=b=c=1$ is the maxima. Thus, the inequality comes.  
