An injective Lipschitz map $f \colon S^{1} \to \mathbb{R}^n$produces a loop $M:=f(S^{1})$.
The inner metric $d_M$ on $M$ is defined by $d_M(x,y):=\inf L(\gamma)$ over all curves $\gamma\subset M$ connecting $x$ with $y$.
It is claimed that
$$\sup\left\{{d_M(x,y)\over \|x-y\|}\>\biggm| x,y\in M, \ x\ne y\right\}\geq{\pi\over2}\ ,$$ whereby $\|x-y\|$ denotes euclidean distance in ${\mathbb R}^n$.
I tried hard resolve this issue and I hope that you can help me. Thank you.
Consider a Lipschitz loop $L\subset {\mathbb R}^n$. By rescaling, it suffices to consider the case of loops of the length $2\pi$. I will parameterize $L$ by its arclength and extend the parameterization to a periodic function $\gamma: {\mathbb R}\to {\mathbb R}^n$. Two points $p, q$ on $L$ are called antipodal if $p=g(t), q=g(t+\pi)$, in other words, $d_L(p,q)=\pi$. The goal is to show that there are antipodal points $p, q$ in $L$ such that $\|p-q\|\le 2$. Define the map $$ f: {\mathbb R}\to {\mathbb R}^n, f(t)= g(t) - g(t+\pi). $$ Then $f$ is $2$-Lipschitz (as a sum of two $1$-Lipschitz functions). Moreover, $f(t+\pi)= - f(t)$ for all $t$. Our claim amounts to that $\|f(t)\|\le 2$ for some $t$. Suppose that such $t$ does not exist, then the image of $f$ is outside of the closed radius 2 ball $B=B(0, 2)\subset {\mathbb R}^n$. Pick an arbitrary $t\in {\mathbb R}$. The path $H=f([t, t+ \pi])$ connects the antipodal points $p=f(t), q=f(t+ \pi)$ and lies outside of the ball $B$. By projecting (using the nearest point projection) this path to the boundary sphere $S$ of the ball $B$, we see that the length of this path is strictly greater than the length of a path in $S$ connecting the antipodal points $$ 2p/\|p\|, 2q/\|q\| \in S. $$ (The nearest point projection to convex solids is $1$-Lipschitz.) Therefore, this length is $\ge 2\pi$ (the length of the spherical semicircle in the sphere of the radius $2$). On the other hand, the path $H$ has length $\le 2\pi$ since $f$ is $2$-Lipschitz. This is a contradiction.
Thus, we proved that $$ \sup_{x\ne y} \frac{d_L(x,y)}{\|x-y\|} \ge \frac{\pi}{2}. $$ qed
