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.
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.
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.
