 Solution

The set $K\cap P$ is compact because $K$ is closed and bounded, and the productis continuous. By the extreme value theorem, $G$ has a maximizer. Since $K$ meets the interior of the nonnegative orthant $P$, some feasible product is positive, so every maximizer lies in $\mathrm{Int}P$.

On $\mathrm{Int}P$, maximizing $G$ is equivalent to maximizingThe Hessian of $F$ iswhich is negative definite. Thus $F$ is a strictly concave function, and it has at most one maximizer on the convex set $K\cap P$. Denote this unique point by $x^{\ast }$.

For any $x\in K\cap P$, the line segmentlies in $K\cap \mathrm{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:ThereforeThis is the supporting inequality for the product maximizer on a compact convex subset of the positive orthant.

Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^{\ast }$. Hencefor some $r>0$. More explicitly,Indeed, cyclic symmetry averaging puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^{\ast }=r$ gives ${\sum }_j{x}_j\le nr$.

Finally, given $a=(a_1,\dots ,a_n)\in \mathrm{Int}P$, defineThis set is nonempty, closed, convex and bounded, and it contains $a$. By the arithmetic-geometric mean inequality,Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has