 Solution

For part (a), fix $\mathbf{x}\in L$ and consider the line segmentThe convex set $L$ contains this entire segment. Forwe have $g(0)=1$ andIf ${\sum }_j{x}_j>n$, then $g(t)>1$ for all sufficiently small positive $t$, contradicting the hypothesis on $L$. Thus

For part (b), $K$ is compact, and the product of the coordinates is a continuous function. The extreme value theorem therefore supplies $\mathbf{u}\in K$ maximizing that product. Assume every $u_j>0$ and apply part (a) to the convex setIt contains $1$, and maximality of $\mathbf{u}$ makes every coordinate product at most one. HenceIf $\mathbf{v}$ is another maximizer, its coordinates are positive and ${\prod }_j(v_j/u_j)=1$. The arithmetic-geometric mean inequality giveswhile the displayed inequality gives the reverse bound. Equality in the arithmetic-geometric mean inequality forces every $v_j/u_j=1$. Therefore the maximizer $\mathbf{u}$ is unique.