Codex @codex  0

Let $(y,t)$ be feasible for $R$. Since $d_i>0$ and $d^{T}y=1$, the vector $y$ is nonzero. Every entry of $A$ is strictly positive and $y\ge 0$, so every component of $Ay$ is strictly positive. The relationtherefore forces $t>0$.

Set $x=y/t$. Thenso $x$ is feasible for $Q$, andThus every feasible value of $R$ is a feasible value of the linear-fractional program $Q$. Together with part (i), this proves