= Solution
Work on the positive orthant, where all real powers in $g$ are defined. To justify a global maximum, enlarge the constraint set to the compact simplex
$$
K=\{x_i\geq0:\sum_i p_ix_i=1\}.
$$
It is closed and bounded since $x_i\leq1/p_i$. The product $g$ extends continuously to $K$, is zero at every boundary point where a coordinate vanishes, and equals $1$ at $(1,\ldots,1)$. Hence a maximum exists and lies in the positive interior. For $n=1$ the constraint gives $x_1=1$ directly.
For $n\geq2$, at an interior maximum the <Lagrange multiplier> equations are
$$
\partial_i g=\frac{p_ig}{x_i}=\lambda p_i.
$$
All $p_i$ and $g$ are positive, so all $x_i$ are equal. The constraint and $\sum_i p_i=1$ make their common value $1$. Thus the only possible maximizer has value $1$, proving
$$
\boxed{g(x)\leq1\quad\text{on }M,}
$$
with equality exactly when every $x_i=1$. The source writes $g$ on all of $\mathbb R^n$; for general noninteger exponents its relevant real-valued domain is the positive orthant used here.
Back to article page
