Hessian is negative definite or negative semidefinite for maxima
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Answer:
Yes, at a local maximum the Hessian of a smooth (real) function will be negative semi-definite (and equivalently the Hessian will be positive semi-definite at a local minimum). If this is hard to find, being a weak converse of the second derivative test, it's likely because proving it requires some linear algebra material not yet covered at the time of a 3rd quarter in calculus where multivariable concepts are introduced.
Consider that the Hessian is a symmetric (real) matrix, and thus has a complete basis of eigenvectors. At a local maximum the function will have, on each line passing through the maximum point, a familiar one-dimensional local maximum. If the Hessian were not negative semi-definite, it would have a line (corresponding to an eigenvector of a positive eigenvalue) along which the restricted function would have a concave up appearance. But this would contradict the point being a local maximum.