Diameter estimate of distance sphere of positive curved manifold Let $M$ be an $n$-dimensional Riemannian manifold with sectional curvature lower bound 1. Fix a point say $O\in M$, let $S(r)$ denote the distance sphere centered at $O$ with radius $r$. The classical Hessian comparison theorem says that the principle curvatures of $S(r)$ is less than that of standard sphere ${S}^n(1)$. And Toponogov triangle comparison implies that given any two point in $S(r)$ there distance in $M$ is less than or equal to the correspond distance in round sphere with the same openning angle at $O$. 
So is there any way to see how the intrinsic diameter (i.e. the length metric induced from ambient metric) upper bound?
How about the Ricci curvature case?
 A: I guess you want to ask is it true that 
$$\mathop{\rm IntrinsicDiameter}[S(r)]\le\mathop{\rm IntrinsicDiameter}[\tilde S(r)],$$
where $\tilde S(r)$ denotes the sphere of radius $r$ in the standard sphere.


*

*This is true if $r\ge \tfrac\pi2$; it follows since $S(r)$ has bigger curvature than $\tilde S(r)$ in the sense of Alexandrov.

*Note that if $r<\tfrac\pi2$ then $S(r)$ might be not connected; in this case 
$$\mathop{\rm IntrinsicDiameter}[S(r)]=\infty.$$ 
If sectional curvature $\ge 1$, I do not see other counterexamples.
It reminds me some questions related to the conjecture that boundary of Alexandrov space is an Alexandrov space. If you find a way to prove it then likely you will get some nontrivial corollaries of this conjecture say if $\Sigma$ is an Alexandrov space with curvature $\ge 1$ then $\mathop{\rm diam}\partial\Sigma\le \pi$ or perimeter of any triangle in $\partial\Sigma$ is at most $2{\cdot}\pi$.
If $r\le\tfrac\pi2$, it is possible to construct a short map $h_r\colon \tilde S(r)\to M$ so that its image covers $S(r)$.
In particular 
$$\text{area}[S(r)]\le\text{area}[\tilde S(r)]$$
(which is obvious anyway).
In general the image of $h_r$ contains creases which stick inside $S(r)$ which in principle might be used as a shortcut. 

*For Ricci curvature the statement does not hold even if $S(r)$ is connected.
You may take a small disc in hyperbolic plane and take a warp product with the sphere to make the Ricci curvature of obtained manifold to be colose to $+\infty$. The sphere $S(r)$ will have intrinsic diameter bigger than $\tilde S(r)$ as far as $S(r)\ne\emptyset$.
