find the min of

From A.M.-G.M. inequality for positive values of $x\,,\,y\,$ and $z$ , we have

$\Large\frac{x + y + z}{3}\,$ $\geq \,\sqrt[3]{xyz}$

$\Rightarrow \Large\frac{(x + y + z)^3}{27}\,$ $\geq\, xyz = x + y + z + 2$

$\Rightarrow (x + y + z)^3 \geq 27(x + y + z + 2)\newline\Rightarrow (x + y + z)^3 - 27(x + y + z) - 54 \geq 0$

Let $x + y + z = c$ . Then

$c^3 - 27c - 54 \geq 0\newline\Rightarrow (c + 3)^2(c - 6) \geq 0\newline \Rightarrow c - 6 \geq 0\newline\Rightarrow c \geq 6\newline \Rightarrow x + y + z \geq 6\newline \Rightarrow xyz - 2 \geq 6\newline \Rightarrow xyz \geq 8$

Now applying A.M.-G.M. inequality on positive numbers $x^2\,,\,y^2\,$ and $\,z^2$

$\Large\frac{x^2 + y^2 + z^2}{3}\,$ $\geq \sqrt[3]{x^2y^2z^2}$

$\Rightarrow x^2 + y^2 + z^2 \,\geq\, 3\sqrt[3]{(xyz)^2} \,\geq\, 3\sqrt[3]{8^2} = 3\sqrt[3]{64} = 3\cdot 4 = 12$

$\Rightarrow x^2 + y^2 + z^2 \geq 12$

Hence $\text{min}(x^2 + y^2 + z^2) = 12$ and equality occurs at $x = y = z = 2$