Consider the following constrained optimization problem:
$$ \min x^3+y^3 \\ \text{s.t.} x^2+y^2\leq 1 $$
From plotting this, the minima seems to be at $(-1,0)$ and $(0,-1)$. Now, the KKT conditions become:
Lagrangian: $L = x^3+y^3+\lambda (-x^2-y^2+1)$ and $\nabla L = 0$
$$=>3x^2-2\lambda x=0$$ $$3y^2-2\lambda y=0$$
Along with $\lambda \geq 0$
When we have $(-1,0)$ the first equation becomes:
$$\lambda = -\frac{3}{2}$$
Since this is contradiction with $\lambda \geq 0$, it seems the KKT conditions are not satisfied here. Isn't that problematic since this does appear to be the minima.
The general framework is this: Let $X \subseteq \mathbb{R}^n$, $f:X\rightarrow\mathbb{R}, g_i:X\rightarrow\mathbb{R}$ for $i \in \{1, ..., k\}$. The optimization problem is: \begin{align} \mbox{Minimize:} \quad & f(x) \\ \mbox{Subject to:} \quad & g_i(x) \leq 0 \quad \forall i \in \{1, ..., k\}\\ & x \in \mathcal{X} \end{align} Now you want to "simultaneously" make $f(x), g_1(x), ..., g_k(x)$ small, so to do that you minimize a weighted sum $$ L(x;\lambda)= f(x) + \sum_{i=1}^k \lambda_i g_i(x)$$ for some multipliers $\lambda_i\geq 0$. It turns out that minimizing $L$ over all $x \in X$ is useful for general (possibly non-differentiable) problems, as is finding stationary points of $L$ for differentiable problems.