= Maximal diameter rigidity from disjoint comparison balls
{title2=$\operatorname{diam}M=\pi/\sqrt k\ \Longrightarrow\ M\cong S_k^n$}
Assume the trace <Ricci curvature> satisfies $\operatorname{Ric}\geq(n-1)kg$, with $k>0$. If points $p,q$ have <Riemannian distance> $D=\pi/\sqrt k$, the balls $B(p,r)$ and $B(q,D-r)$ are disjoint. <Bishop volume comparison with a positive-curvature model> bounds their volumes below by $\operatorname{Vol}(M)v_k(r)/v_k(D)$ and $\operatorname{Vol}(M)v_k(D-r)/v_k(D)$. Since $v_k(r)+v_k(D-r)=v_k(D)$, both inequalities must be equalities. Taking $r\downarrow0$ gives $\operatorname{Vol}M=v_k(D)$, and <spherical rigidity of maximal total volume> yields the round sphere.
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