algebra precalculus - Interesting short Inequality - Mathematics Stack Exchange

this problem romanian g.m. , although short, (i think) hard. let $x, y, z$ real non-negative numbers such $x+y+z=3$. prove

$$27 \leq (x^2+2)(y^2+2)(z^2+2) \leq 44 $$

i didn't succeed in of inequalities, managed find equality cases ($x=y=z=1$ first 1 , 1 of them $3$ , others $0$). hint/idea/solution welcome.

for proof of left inequality it's enough prove $$(x^2+2)(y^2+2)(z^2+2)\geq3(x+y+z)^2,$$ true reals $x$, $y$ , $z$.

the proof see here: contest inequality - gm?

for proof of right inequality it's enough prove $$44(x+y+z)^6\geq(2(x+y+z)^2+9x^2)(2(x+y+z)^2+9y^2)(2(x+y+z)^2+9z^2),$$ $$\sum_{sym}\left(4x^5y+7x^4y^2+3x^3y^3+18x^4yz+70x^3y^2z+\frac{51}{4}x^2y^2z^2\right)\geq0,$$ obvious.