verifying extrema found by Lagrange multipliers

Question

This question was inspired by reading this problem: Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$

Suppose I have a function $f(x,y,z)$ with continuous partial derivatives which I want to maximize and minimize on the portion of the plane $x+y+z=k$ with $x,y,z>0$.

If I use Lagrange multipliers and find the points $(a_i,b_i,c_i)$ for $1\le i\le4$ where extrema could occur

[by solving the system $f_x=\lambda\cdot1, f_y=\lambda\cdot1, f_z=\lambda\cdot1$],

and $f(a_1,b_1,c_1)=2$ and $f(a_2,b_2,c_2)=f(a_3,b_3,c_3)=f(a_4,b_4,c_4)=\frac{7}{4}$,

what additional information, if any, do I need in order to conclude that 2 is the maximum value of $f$ or

that $\frac{7}{4}$ is the minimum value of $f$ on the set of points being considered?

Answer 1

If $f$ is smooth enough to have second partial derivatives, you could use the second derivative test for constrained problems. You do this by forming the bordered Hessian,

\begin{equation} H(x, y, z) = \left[\begin{array}{ccccc} 0 & \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \\ \frac{\partial g}{\partial x} & \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial g}{\partial y} & \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial g}{\partial z} & \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \\ \end{array}\right]. \end{equation}

To determine if a point $(x_0, y_0, z_0)$ is a minimum or a maximum, we plug that point into $H$ and then look at $n - m$ of the bordered Hessian's principal minors where $n$ is the number of variables and $m$ is the number of constraints. Here, then, we would look at two and this makes sense because we have $3$ variables and one constraint, so in a sense we have two variables which are "free". First we examine the minor made up of the first $2m+1 = 3$ rows and columns of $H$ and compute its determinant. Then we look at the minor made up of the first $2m + 2 = 4$ rows and columns and compute its determinant; this second determinant is just the determinant of $H(x_0, y_0, z_0)$ itself.

For this problem, a sufficient condition for a local minimum of $f$ is to have both of the determinants we computed above be negative. A sufficient condition for a local maximum of $f$ for this problem is that the determinant of the first minor have be positive and that the determinant of the second minor be negative. In the event that these conditions aren't met, this test doesn't tell us anything useful.