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The active constraint set is $\boxed{\mathcal{A}(x)=\{j:c_j(x)=0\}}$, and the linearized tangent cone isDifferentiating each active feasibility inequality along a tangent sequence gives $T_{\Omega }(x)\subseteq F(x)$. A sufficient condition for equality is the linear independence constraint qualification: the active constraint gradients are linearly independent. The weaker Mangasarian–Fromovitz constraint qualification also suffices in this inequality-only setting.

Under the stated independence condition, the necessary first-order conditions become the Karush-Kuhn-Tucker conditions: there are multipliers ${\lambda }_j$ withUse the allowed span assertion to write $\nabla f(x_{\ast })={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast })$. Independence gives vectors ${\phi }_i$ with $\nabla c_j(x_{\ast }{)}^T{\phi }_i={\delta }_{ji}$ for the active indices. Each ${\phi }_i$ lies in $F=T_{\Omega }$, so part (a) gives ${\lambda }_i=\nabla f(x_{\ast }{)}^T{\phi }_i\ge 0$. Assign zero multipliers to inactive constraints. This proves all displayed conditions, including complementarity. Conversely, their stationarity and sign conditions imply $\nabla f(x_{\ast }{)}^{T}d={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast }{)}^{T}d\ge 0$ for every $d\in F$. Thus they reformulate the first-order inequality; they do not alone prove a minimum for a nonconvex problem.