Let $p,q,r$ be fixed positive constants.
Edited to include the constraint:
The constraint is $g(x,y,z)=xyz=C$, where $C$ is positive.
I am trying to show that the one critical point that I found, for the function
$$f(x,y,z)=\frac{x^p}{p}+\frac{y^q}{q}+\frac{z^r}{r}$$
is a minimizer, so I think I would want to show that the Hessian matrix of $f$, evaluated at the critical point, is positive-definite.
The Hessian was computed easily enough -- it's diagonal, so the eigenvalues of the matrix can be just be read off.
However, the diagonal elements are of the form $(p-1)C^A$, ... , $(r-1)C^D$,
where $C$ is positive.
Although we know that $p,q,r$ are positive numbers by assumption, we don't know whether any of $(p-1)$, $(q-1)$, or $(r-1)$ are positive or negative.
So, I can't say for sure that the matrix has positive eigenvalues.
Should I instead use another approach to convince myself that this one critical point found must be a minimizer?
When exponents get in the way, logarithms come to the rescue. Introduce new variables $u=\log x$, $v=\log y$, $w=\log z$. The problem changes to minimizing $$f(u,v,w)=\frac{1}{p}e^{pu}+\frac{1}{q}e^{qv}+\frac{1}{r}e^{rw}$$ subject to the linear constraint $u+v+w=\log C$.
The linearity of constraint is important, because it simplifies the second derivative test for constrained extremum. With a linear constraint, we just restrict attention to a subspace when testing the positive/negative definiteness of the Hessian. With a nonlinear constraint we enter the world of bordered Hessians — the ordinary Hessian no longer suffices, because the curvature of the surface to which the function is constrained can affect the nature of extremum.
In the above case, $f$ is obviously convex (sum of convex functions), so its restriction to a hyperplane is convex, too.