= Solution
For <constrained optimization> in minimization form, let $x\in X$, $g_i(x)\le0$ and $h_j(x)=0$, and define the <Lagrangian>
$$
L(x,\lambda,\nu)=f(x)+\sum_i\lambda_i g_i(x)+\sum_j\nu_jh_j(x).
$$
The <Lagrangian sufficiency theorem> says: if $x_*$ is feasible, $\lambda_i\ge0$, $\lambda_i g_i(x_*)=0$ for every $i$, and $x_*$ minimizes $L(\cdot,\lambda,\nu)$ globally over $X$, then $x_*$ minimizes $f$ over the feasible set. Equality multipliers $\nu_j$ may have either sign.
Indeed, for every feasible $x$,
$$
f(x)\ge L(x,\lambda,\nu)\ge L(x_*,\lambda,\nu)=f(x_*).
$$
The first inequality uses the inequality-multiplier signs and the equality constraints; the last equality uses <complementary slackness>. This proves global optimality without assuming <convexity> or differentiability. The maximization version reverses the relevant signs, or follows by replacing $f$ with $-f$.
When $X$ is convex and the Lagrangian is differentiable and convex, the first-order condition $\nabla_xL(x_*)=0$ implies its global minimum, because $L(x)\ge L(x_*)+\nabla L(x_*)\cdot(x-x_*)$. Thus feasible stationarity plus <complementary slackness> is sufficient in that convex setting. Mere stationarity in a general nonconvex problem is not the theorem's global-minimum hypothesis.
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