Find the minimum of $(x(1+y)+y(1+z)+z(1+x))/\sqrt{xyz}$ over positive integers $x,y,z$ Let $x,y,z$ be positive integers.The least value 
$$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{\frac 12}}$$
I tried sum using arithmetic-geometric means inequality (seems promising as the denominator is similar to geometric mean of $x,y,z$). However, so far this didn't help my cause.
 A: Notice
$$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{\frac 12}} = \frac{x + xy + y + yz + z + zx}{(xyz)^{1/2}} \geq \frac{ 6 (x^3y^3z^3)^{1/6}}{(xyz)^{1/2}} = 6$$
By the AM-GM inequality
A: Bryansis2010's answer is spot on. You take Partial Derivatives of your expression with respect to each variable. Note that a partial derivative says to take the derivative of the function with respect to a single variable, treating the others as constant. So for example, given $f(x, y) = x^{2}y$, $\dfrac{\partial f}{\partial x} = 2xy$ (this is the first partial derivative of $f$ with respect to $x$), while $\dfrac{\partial f}{\partial y} = x^{2}$. All your derivative rules from Calc I hold.
So Bryansis2010's solution gives you candidates to check. If a function attains its maximum or minimum, then the partial derivatives at that point are all $0$. So you find points such that the partial derivatives are $0$, then you check to see which of them are minimizers.
It looks like you've gotten it figured out, but I leave this explanation as an alternative and for others who may view this thread.
A: Take partial derivatives and then solve for them = $0$.
