 Solution

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 $\log (xyz)$.