A nice use

Since $x, y, z$ are positive, we can use AM-GM inequality.

$\dfrac{x^2+2xy+2xy+4y^2+z^2+z^2}{6}\geq (x^2×2xy×2xy×4y^2×z^2×z^2)^{\frac{1}{6}}$

$\Rightarrow x^2+4xy+4y^2+2z^2\geq (16x^4y^4z^4)^{\frac{1}{6}}=\boxed{96}$

The objective behind splitting the terms was to get something like $(xyz)^n$ in product side.

x^2+4xy+(4y^2)+(2z^2)=x^2+2xy+2xy+4y^2+z^2+z^2=k Using AM-GM inequality on these 6 terms, ( k/6)>=(16*(xyz)^4)^(1/6)=16 Hence, k>=16x6=96