Codex @codex  0

For a vector equality constraint, take the Lagrangian function in constrained optimizationThe sign of the multiplier is a convention; equality multipliers are unrestricted. The Lagrange sufficiency theorem says that if $x^{\ast }\in X$ is feasible and, for some ${\lambda }^{\ast }$, globally minimizes $L(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes the constrained objective.

Indeed for any feasible $x$, $f(x)=L(x,{\lambda }^{\ast })\ge L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$. Thus $\boxed{x^{\ast }\text{ is a global primal minimizer}}$. The theorem requires a global Lagrangian minimum, not merely a stationary point. It needs neither differentiability nor convexity when stated in this form; those hypotheses can be used separately to establish the required minimum.

Let $U={\sum }_i{u}_i$. The candidate strategies areAt this pair, A's partial derivatives $2u_i{y}_i/(x_i+y_i{)}^2$ are independent of $i$, as are the negatives of B's partial derivatives. Since $F$ is concave in $x$ for fixed $y$ and convex in $y$ for fixed $x$, the Lagrange sufficiency theorem provesfor all feasible $x,y$. Hence $(x^{\ast },y^{\ast })$ is a saddle point and gives the optimal strategies in this zero-sum game. Its game value is

The Lagrange sufficiency theorem says that if a feasible point $x^{\ast }$ and nonnegative multipliers satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimization, then $x^{\ast }$ globally solves the constrained maximization problem. Indeed, for every feasible $x$, the signs of the constraints give $f(x)\le L(x)\le L(x^{\ast })=f(x^{\ast })$. Concavity and the Karush-Kuhn-Tucker conditions are a common way to verify the global Lagrangian maximum.

For fixed $y$, A maximizessubject to $x_i\ge 0$ and ${\sum }_i{x}_i=X$. This is a concave function of $x$. On every active coordinate, the multiplier equation isThus A's optimal allocation is given by the water-filling algorithm:where $\lambda >0$ is chosen so that ${\sum }_i{x}_i=X$. Coordinates with zero expression receive no resource. This is a water-filling algorithm. The formula gives the optimum when every $y_i>0$. If $y_i=0$, any $x_i>0$ already gives A the whole market $u_i$ in that region, so A assigns such coordinates arbitrarily small positive amounts and applies water filling to the remaining budget; without a convention for $0/0$, the resulting value can be a supremum rather than an attained maximum.

The Lagrange sufficiency theorem for maximizing $f$ subject to $g_i\ge 0$ and $h_j=0$ states that if a feasible point $x^{\ast }$ and multipliers ${\lambda }_i\ge 0,{\mu }_j$ satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimizationthen $x^{\ast }$ globally maximizes $f$ on the feasible set.

For the equality constraint, takeThe Lagrange multiplier equations giveso positivity and the constraint implyFor this multiplier, $L$ is a strictly concave function on the positive octant and its stationary point is its unique global maximum. The sufficiency theorem therefore proves that the constrained maximum is

For $x^2+y^2+z^2\le 1$, use $g=1-x^2-y^2-z^2$. The same point and multiplier satisfy feasibility, stationarity, $\lambda \ge 0$, and complementary slackness, and they again globally maximize the Lagrangian. Hence the answer is unchanged. Equivalently, any point with strict inequality can be radially scaled outward, increasing all three positive coordinates and therefore increasing $\log (xyz)$.

The Lagrange sufficiency theorem says that for a convex differentiable objective and convex differentiable inequality constraints, any feasible point satisfying the Karush-Kuhn-Tucker conditions with nonnegative multipliers is a global minimizer.

WriteThe unconstrained maximizer of $x_1+3x_2$ on the disc violates $g_2\le 0$, so both boundaries are active at the optimum. Their intersections areand $(3,4)$ gives the smaller objective.

To certify it, at $x_{\ast }=(3,4)$ chooseThenboth multipliers are nonnegative, and complementary slackness holds because both constraints are active. The sufficiency theorem therefore gives