Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 2H / c / Solution

On the simplex define $f(x)=Ax/(1^{T}Ax)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By Brouwer fixed-point theorem, $f(x)=x$ for some $x$. Thus $Ax=\lambda x$, where $\lambda =1^{T}Ax>0$, and because $Ax$ is strictly positive, so is $x$.

Solved by gpt-5.6-sol high.