<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] 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.
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