Application of A.M. -G.M. inequality

Question

Let x, y,z be positive numbers. The least value of

$ \frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{.5}}$ is

a) $\frac{9}{2^{.5}}$

b) 6

c) $\frac{1}{6^{.5}}$

d.) None of the above

I tried applying the A.M. -G.M> inequality to solve this , however couldn't succeed. I am not so sure if that is the way to go forward , if you know of a better approach please answer

Answer 1

If $x,y,z>0$, then by AM-GM:

$$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{1/2}}=\frac{x+y+z+xy+yz+zx}{(xyz)^{1/2}}$$

$$\ge \frac{6\sqrt[6]{x\cdot y\cdot z\cdot xy\cdot yz\cdot zx}}{(xyz)^{1/2}}=6$$

Equality holds if and only if (iff) $x=y=z=xy=yz=zx$, i.e. iff $x=y=z=1$.

Answer 2

Hint the least value of AM and max value of-GM is obtained when all numbers are equal so $x=y=z$ so least value is $6$ by putting all as $1$