Proof. Necessity (sketch). At a local minimum there is no direction \(d\) that decreases \(f\) to first order, \(\nabla f^{\mathsf T}d < 0\), while keeping the active inequality constraints, \(\nabla g_i^{\mathsf T}d \leq 0\), and the equalities, \(\nabla h_k^{\mathsf T}d = 0\), satisfied to first order; the constraint qualification makes this linearisation accurate. By Farkas’ lemma (Theorem 5.4.1), the absence of such a \(d\) is equivalent to the existence of the multipliers, with \(\lambda _i = 0\) for the inactive constraints.

Sufficiency. Let \(x\) be feasible. By convexity, \(f(x) \geq f(x^*) + \nabla f(x^*)^{\mathsf T}(x - x^*)\), and by stationarity \(\nabla f(x^*)^{\mathsf T}(x - x^*) = -\sum _i\lambda _i\nabla g_i(x^*)^{\mathsf T}(x - x^*) - \sum _k\mu _k\nabla h_k(x^*)^{\mathsf T}(x - x^*)\). The last sum is \(0\) since \(h_k\) is affine and \(h_k(x) = h_k(x^*) = 0\). For each \(i\), convexity of \(g_i\) gives \(\nabla g_i(x^*)^{\mathsf T}(x - x^*) \leq g_i(x) - g_i(x^*)\), and \(\lambda _i\bigl (g_i(x) - g_i(x^*)\bigr ) = \lambda _ig_i(x) \leq 0\) by complementary slackness and feasibility. So \(f(x) \geq f(x^*)\). □