Strictness threshold in the \xi_n–\theta_n inequality for the cube

Prove or refute that the smallest dimension n for which the inequality \xi_n(Q_n) < \frac{n+1}{2}(\theta_n(Q_n) - 1) + 1 holds (i.e., the inequality is strict) is n=4.

Background

For all n, the inequality \xi_n(Q_n) ≤ \frac{n+1}{2}(\theta_n(Q_n) - 1) + 1 holds. Equality is known for n=1,2,3,7, the same cases where both \theta_n(Q_n) and \xi_n(Q_n) are exactly known.

The conjecture pinpoints n=4 as the first dimension where strict inequality should hold, but this has not been settled.