Interesting short Inequality This problem is from Romanian G.M. and although it is very short, it is also (I think) very hard.
Let $x, y, z$ real non-negative numbers such that $x+y+z=3$. Prove that 
$$27 \leq (x^2+2)(y^2+2)(z^2+2) \leq 44 $$
I didn't succeed in any of the inequalities, I only managed to find the equality cases ($x=y=z=1$ for the first one and one of them $3$ and the others $0$). Any hint/idea/solution is welcome.  
 A: Try the following method. You are looking for the extrema of the function
$f\left(x,y,z\right)=\left(x^{2}+2\right)\left(y^{2}+2\right)\left(z^{2}+2\right)$
under the constraint $g\left(x,y,z\right)=x+y+z-3=0$. Put $D=\left\{ \left(x,y,z\right)\in\left(\mathbb{R}_{+}\right)^{3}:x+y+z-3=0\right\} $. 
You can look after the extrema of $f$ in the interior of $D$ by
searching the points $a\in\overset{\circ}{D}$ for which the gradients
of $f$ and $g$ are colinear (Lagrange's method). Then look after
the extrema of $f$ on the boundary of $D$, which is obtained by
letting one of the variables $x,y$ or $z$ equal to $0$.
Then compute $f$ at all these points to detect its minima and maxima.
A: For the proof of the left inequality it's enough to prove that
$$(x^2+2)(y^2+2)(z^2+2)\geq3(x+y+z)^2,$$
which is true even for all reals $x$, $y$ and $z$.
The proof see here: Contest Inequality - Is it AM GM?
For the proof of the right inequality it's enough to prove that
$$44(x+y+z)^6\geq(2(x+y+z)^2+9x^2)(2(x+y+z)^2+9y^2)(2(x+y+z)^2+9z^2),$$
which is
$$\sum_{sym}\left(4x^5y+7x^4y^2+3x^3y^3+18x^4yz+70x^3y^2z+\frac{51}{4}x^2y^2z^2\right)\geq0,$$
which is obvious.
