The two forms of $\omega_{AB}$ found in part (i) give
$$
(1+\varepsilon)\omega_{AB}-\rho_{AB}=X_-\geq0,
\qquad
(1+\varepsilon)\omega_{AB}-\sigma_{AB}=X_+\geq0.
$$
Therefore, in the <Löwner order>,
$$
\boxed{\rho_{AB}\leq(1+\varepsilon)\omega_{AB}},
\qquad
\boxed{\sigma_{AB}\leq(1+\varepsilon)\omega_{AB}}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 4 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-323/4/c/ii/solution