I am going to post a particular example for the sake of clarity.

One needs to maximize a real function

$F = a_1a_2 + a_2a_3 + \cdots + a_{n - > 1}a_n + a_na_1;$

with active constraints

$h_1 = a_1 + a_2 + \cdots + a_n = 0;$

$h_2 = a_1^2 + a_2^2 + \cdots +a_n^2 = > 1;$

The gradients in question

$\nabla F = (a_2+a_n,a_3+a_1, \ldots , a_{n-2}+a_n,a_{n-1}+a_1);$

$\nabla h_1 = (1, 1, \ldots ,1, 1);$

$\nabla h_2 = (2a_1,2a_2,\ldots , 2a_{n-1}, 2a_n);$

Because of Kuhn–Tucker conditions it is reasonable to write $\nabla F$ in terms of $\nabla h_1$ and $\nabla h_1$

$$\begin{cases} a_2+a_n = \lambda_1 + 2\lambda_2a_1; \\\ a_3+a_1 = \lambda_1 + 2\lambda_2a_2; \\\ a_4+a_2 = \lambda_1 + 2\lambda_2a_3; \\\ \ldots \\\ a_{n-2} + a_n = \lambda_1 + 2\lambda_2 a_{n-1}; \\\ a_{n-1} + a_1 = \lambda_1 + 2\lambda_2 a_n; \end{cases}$$

$$\begin{pmatrix} 2\lambda_2 & -1 & 0 & 0 & \ldots & 0 & -1 \\\ -1 & 2\lambda_2 & -1 & 0 & \ldots & 0 & 0 \\\ 0 & -1 & 2\lambda_2 & -1 & \ldots &0 & 0 \\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\\ 0 &0 &0 & 0 & \ldots & 2\lambda_2 & -1 \\\ -1 &0 &0 & 0 & \ldots & -1 & 2\lambda_2 \end{pmatrix} \begin{pmatrix} a_1 \\\ a_2 \\\ a_3 \\\ \vdots \\\ a_{n-1} \\\ a_n \end{pmatrix} = - \lambda_1\begin{pmatrix} 1 \\\ 1 \\\ 1 \\\ \vdots \\\ 1 \\\ 1 \end{pmatrix} $$

By adding up all equations it follows that $\lambda_1 = 0$ so it's a matter of nullspaces.

However is this attempt fruitless? I'm getting nowhere, so at this point references to similar problems would also be appreciated.

A few remarks:

- Case $n=2$ is somewhat trivial, for all valid combinations (there are two of them) $F = -1$.
- Case $n=3$ brings up the same problem since for all valid combinations $F = -1/2$.
- $F \le 1$ is an exact upper bound (it follows from properties of scalar multiplication and constructing silly yet valid combinations for sufficiently large $n$).