The minimum of $\dfrac{x(1+y)+y(1+z)+z(1+x)}{\sqrt{xyz}}?$

Question

Can we use $AM\geq GM$ inequality to find the minimum of $\dfrac{x(1+y)+y(1+z)+z(1+x)}{\sqrt{xyz}}?$

I can find out that minimum is $6$, but can we use $AM\geq GM$ to show this?

Answer 1

Expand the expression $$x(1+y)+y(1+z)+z(1+x) = x+yz+y+xz+z+yx $$ From $\rm{AM} \geq \rm{GM}$ we get $ x+yz \geq 2 \sqrt{xyz}$, do the same for the other terms and add to conclude.

Answer 2

Alternately: Using AM-GM for $6$ positive numbers: $ x + y + z + xy + yz + zx \geq 6\sqrt[6]{x\cdot y\cdot z\cdot xy\cdot yz\cdot zx} = 6\sqrt[6]{x^3\cdot y^3\cdot z^3} = 6\sqrt{xyz}$. So $\dfrac{x+y+z+xy+yz+zx}{\sqrt{xyz}} \geq 6$ with equality at $x = y = z = xy = yz = zx$ implying $x = y = z = 1$