 Solution

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.