The Other AM-GM (Part 1)

By AM-GM, we have $x+ 2y \geq 2\sqrt{2xy}$ ; $y+2z \geq 2\sqrt{2yz}$ ; $xz+ 1 \geq 2\sqrt{xz}$ .

Multiply those three inequalities, we have

$(x+2y)(y+2z)(xz+1) \geq 8\sqrt{4x^2y^2z^2}$

$(x+2y)(y+2z)(xz+1) \geq 16 \sqrt{(xyz)^2} = 16 xyz = 16$

So, the minimum value of the given expression $(x+2y)(y+2z)(xz+1)$ is $16$ , when $x= 2, \space y=1, \space z= \frac{1}{2}$ .

Then, we have $x+y+z= \frac{7}{2} = \frac{m}{n}$ .

Hence, $m+n = \boxed{9}$ .

$\color{#3D99F6}{(x+2y)}\color{#D61F06}{(y+2z)}\color{#20A900}{(xz+1)} \stackrel{\color{#333333}{\text{AM-GM}}}\geq \color{#3D99F6}{(2\sqrt2\sqrt{xy})}\color{#D61F06}{(2\sqrt2\sqrt{yz})}\color{#20A900}{(2\sqrt{xz})}\\ \implies (x+2y)(y+2z)(xz+1)\geq 16\sqrt{(x^2y^2z^2)}=16$

Equality occurs when $x=2y,y=2z,xz=1\implies x=\dfrac{1}{z}=4z$ ,hence $z=\dfrac{1}{2}$ from where we get $y=1,x=2$ .Hence $x+y+z=2+1+\dfrac{1}{2}=\dfrac{7}{2}\implies 7+2=\boxed{9}$