= Solution
For part (a), fix $\mathbf x\in L$ and consider the line segment
$$
\mathbf y(t)=(1-t)\mathbf1+t\mathbf x,
\qquad 0\leq t\leq1.
$$
The <convex set> $L$ contains this entire segment. For
$$
g(t)=\prod_{j=1}^n\bigl(1+t(x_j-1)\bigr)
$$
we have $g(0)=1$ and
$$
g'(0)=\sum_{j=1}^n(x_j-1)=\sum_{j=1}^nx_j-n.
$$
If $\sum_jx_j>n$, then $g(t)>1$ for all sufficiently small positive $t$, contradicting the hypothesis on $L$. Thus
$$
\boxed{\sum_{j=1}^nx_j\leq n.}
$$
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 set
$$
L=\left\{\left(\frac{x_1}{u_1},\ldots,\frac{x_n}{u_n}\right):\mathbf x\in K\right\}.
$$
It contains $\mathbf1$, and maximality of $\mathbf u$ makes every coordinate product at most one. Hence
$$
\boxed{\sum_{j=1}^n\frac{x_j}{u_j}\leq n\qquad(\mathbf x\in K).}
$$
If $\mathbf v$ is another maximizer, its coordinates are positive and $\prod_j(v_j/u_j)=1$. The <arithmetic-geometric mean inequality> gives
$$
\frac1n\sum_j\frac{v_j}{u_j}\geq
\left(\prod_j\frac{v_j}{u_j}\right)^{1/n}=1,
$$
while the displayed inequality gives the reverse bound. Equality in the arithmetic-geometric mean inequality forces every $v_j/u_j=1$. Therefore \b[the maximizer $\mathbf u$ is unique].
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