How do I find the infimum of a function using Calculus III techniques? > Let $f(x, y, z) = \frac{x + y + z}{2} - \sqrt{xyz}$. Find the infimum of the function where the domain is restricted to the first quadrant. There are techniques in Calculus III involving the hessian and partial derivatives to calculate the minimum of a function on a compact set. Because the definition of an infimum is the minimum of f except the minimum doesn't have to be in the image f(A), there should be something _similar_ to these methods for calculating the infimum. But here's the provisio: I want the infimum to be calculated **using the methods of Calculus III**. That is, I do not want to use the arithmetic-geometric mean inequality, even though that tells you what the infimum must be.

Take all the partials:
∂f/∂x = 1/2 - √(yz/x)/2 = 0
∂f/∂y = 1/2 - √(xz/y)/2 = 0
∂f/∂z = 1/2 - √(xy/z)/2 = 0
solving these yields the following:
1 = √(yz/x) thus yz=x
1 = √(zx/y) thus zx=y
1 = √(yx/z) thus xy=z

multiplying all these equations yields: (xyz)^2 = (xyz)

thus either xyz = 1 or xyz = 0
say xyz=0, then one of x y or z must be 0
say for the moment its x, then we have
y=zx=z*0=0 and z=xy=0*y=0 which give us that x=y=z=0
similarly if we assume y=0 or z=0 we will get x=y=z=0
thus the case xyz=0 reduces to x=y=z=0

we can now look at the xyz=1 case, notice none of x,y,z can be 0
substituing y=zx into xy=z we have xzx=z
thus because z≠0 we have x^2=1 thus x=1 because x>0
by the symmetry of the conditions we can conclude y=1 and z=1
thus the xyz=1 case reduces to x=y=z=1

so we have 2 potential solutions: (x,y,z) = (0,0,0) or (1,1,1)
substituting them both into f(x,y,z) gives 0 in both cases
so we can conclude they are both the mins