If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is 
$\bf{My\; Try::}$ Here I have Used $\bf{A.M\geq G.M}$ Inequality
So $$\displaystyle \frac{x^2+4xy+4y^2+4z^2}{4}\geq \left(x^3\cdot y^3\cdot z^2\right)^{\frac{1}{4}}\;,$$ But I did not How can I solve it
Help me, Thanks 

Applying the AM-GM is the right strategy, but you need to do it a bit differently. 
$$x^2+4xy+4y^2+4z^2$$ 
$$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$ 
$$\ge 6\sqrt[6]{x^2 \cdot 2xy \cdot 2xy \cdot 4y^2 \cdot 2z^2 \cdot 2z^2}$$ 
$$= 6\sqrt[6]{64x^4y^4z^4}$$ 
$$= 12(xyz)^{2/3}$$ 
$$= 12 \cdot 32^{2/3}$$
Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$. 
