The primal problem is
\begin{align}
\operatorname{minimize}_x & \quad f_0(x) \\
\text{subject to} & \quad f_i(x) \leq 0 \quad \text{for } i = 1,\ldots, m\\
& \quad Ax = b.
\end{align}
The functions $f_i, i = 0,\ldots,m$ are differentiable and convex.
Assume $x^*$ is feasible for the primal problem (so $f_i(x^*) \leq 0$ for $i = 1,\ldots,m$ and $A x^* = b$) and that there exist vectors $\lambda \geq 0$ and $\eta$ such that
$$
\tag{$\spadesuit$}\nabla f_0(x^*) + \sum_{i=1}^m \lambda_i \nabla f_i(x^*) + A^T \eta = 0
$$
and
$$
\lambda_i f_i(x_i) = 0 \quad \text{for } i = 1,\ldots,m.
$$
Because the functions $f_i$ are convex, equation ($\spadesuit$) implies that $x^*$ is a minimizer (with respect to $x$) of the Lagrangian
$$
L(x,\lambda,\eta) = f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \eta^T(Ax - b).
$$
Thus, if $x$ is feasible for the primal problem, then
\begin{align*}
f_0(x) & \geq f_0(x) + \sum_{i=1}^m \lambda_i f_i(x) + \eta^T(Ax - b) \\
& \geq f_0(x^*) + \sum_{i=1}^m \lambda_i f_i(x^*) + \eta^T(Ax^* - b) \\
& = f_0(x^*).
\end{align*}
This shows that $x^*$ is a minimizer for the primal problem.
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 inequalities cannot be taken into account in KKT conditions.
To solve the problem, compute stationary points of the problem without these constraints, then throw away all of them that violate these conditions.