A problem could be
$\texttt{max} \ f(x,y)=-(x-1)^2-(y-1)^2$ under the constraints $x+y\leq 1$ and $x,y\geq 0.$
The lagrange function then is:
$L(x,y,\lambda )=-(x-1)^2-(y-1)^2+\lambda (1-x-y)$
The expression in the brackets of $\lambda ()$ has to be greater or equal to zero.
The KKT conditions are:
$\frac{\partial L}{\partial x}=-2(x-1)-\lambda\leq 0 \quad (1), \quad
\frac{\partial L}{\partial y}=-2(y-1)-\lambda \leq 0 \quad (2)$
$\frac{\partial L}{\partial \lambda}=1-x-y\leq 0\quad (3), \quad
x\cdot \frac{\partial L}{\partial x}=-x\left( 2(x-1)+\lambda\right)= 0\quad (4)$
$y\cdot \frac{\partial L}{\partial y}=-y(2(y-1)+\lambda) = 0 \quad (5), \quad \lambda\cdot \frac{\partial L}{\partial \lambda}=\lambda(1-x-y)=0 \quad (6), \quad x,y,\lambda \geq 0 \quad (7)$
Now you check the two cases $\lambda=0$ and $\lambda \neq 0 $
For (4) and (5) you would have 4 possible solutions $(x,y,\lambda)$:$(0,0,0),(1,0,0),(0,1,0),(1,1,0) $
None of these solutions satisfies the conditions (1),(2) and (3) simultaneously.
Because of (6) we have $1-x-y=0$
If $x=0$, then $y=1$. Inserting the values in (5): $-1(0-\lambda)=\lambda=0$ Because of $\lambda \neq 0$ (case 2) we have a contradiction.
If $x=1$, then $y=0$. Inserting the values in (4): $-1(0-\lambda)=\lambda=0$ Because of $\lambda \neq 0$ (case 2) we have a contradiction.
We can conclude, that $x,y \neq 0$. Because of (4) and (5) we have the two equations.
$2x-2-\lambda=0$
$2y-2-\lambda=0$
Substracting the second equation by the first equation.
$2x-2y=0 \Rightarrow 2x=2y \Rightarrow x=y$
With (6) and $\lambda\neq 0$ we get $1-x-x=0 \Rightarrow 1=2x \Rightarrow x=\frac{1}{2}$ and $y=\frac{1}{2}$ By using (5) we can calculate $\lambda$.
$2\cdot \left(\frac{1}{2}-1\right)+\lambda=0\Rightarrow \lambda=1$
Best Answer
Strict inequality constraints are never binding at a feasible solution (if they were, they wouldn't be strict inequalities). However, they can cause the problem to have no global maximum. What you might do is solve the problem with the strict inequalities replaced by non-strict inequalities. If this new problem has a global maximum where those strict inequalities are satisfied, then that is a global maximum of the original problem. If not, the original problem has no global maximum.
EDIT: Hmm, that's not quite always correct: it is true if the feasible region of the original problem is dense in the feasible region of the new problem. It is true if the new problem has a global maximum $p$ such that the gradients of the strict inequality constraints that are binding at $p$ are linearly independent.