Let's fix a triangle in the plane:
$\Delta=\{t_1x_1+t_2x_2+t_3x_3 \in \mathbb{R}^2 \mid t_1,t_2,t_3 \ge 0 \quad \land \quad t_1+t_2+t_3=1\}$
of vertices $x_1,x_2,x_3 \in \mathbb{R}^2$, and an open disk:
$D=\{x \in \mathbb{R}^2 \mid \|x-x_0\|<r\}$
of radius $r>0$ and center $x_0 \in \Delta$.
Suppose that the perimeter of $\Delta$ is inferior to $2r$, namely:
$\|x_2-x_1\|+\|x_3-x_2\|+\|x_1-x_3\|<2r$.
I want to show that $\Delta \subseteq D$.
Here is my attempt:
$x_0 \in \Delta \quad \Rightarrow \quad x_0=t_1x_1+t_2x_2+t_3x_3$,
with $t_1+t_2+t_3=1$ and $t_1,t_2,t_3 \ge 0$ (and so $t_1,t_2,t_3 \in [0,1]$).
We have $\|x_1-x_0\|=\|x_1-t_1x_1-t_2x_2-t_3x_3\|=\|(1-t_1)x_1-t_2x_2-t_3x_3\|=$
$=\|(t_2+t_3)x_1-t_2x_2-t_3x_3\| \le t_2\|x_1-x_2\|+t_3\|x_1-x_3\| \le$
$\le \|x_1-x_2\|+\|x_1-x_3\|<2r-\|x_3-x_2\|$.
Here I get stuck. If I could show that $\|x_1-x_0\|<r$ (and similarly $\|x_2-x_0\|<r,\|x_3-x_0\|<r$), then I could easily show that $\Delta \subseteq D$.
Thank you!
First notice that from $\|x_2-x_1\|+\|x_3-x_2\|+\|x_1-x_3\|<2r$ follows that $$\max\{\|x_2-x_1\|,\|x_3-x_2\|,\|x_1-x_3\|\} < r.$$ Indeed, if e.g. $\|x_2-x_1\| \ge r$ then by triangle inequality we would get $$r \le \|x_2-x_1\| \le \|x_2-x_3\|+ \|x_1-x_3\|$$ so altogether $\|x_2-x_1\| +\|x_2-x_3\|+ \|x_1-x_3\| \ge 2r$ which is a contradiction.
Write $x_0 = t_1x_1+t_2x_2+t_3x_3$ so \begin{align} \|x_0-x_1\| &= \|t_1x_1 + t_2x_2+t_3x_3 -(t_1 + t_2+t_3)x_1\|\\ &\le t_2\|x_2-x_1\| + t_3\|x_3-x_1\|\\ &\le (t_1+t_2)\max\{\|x_2-x_1\|,\|x_3-x_2\|,\|x_1-x_3\|\}\\ &< r \end{align} so $x_1 \in B(x_0,r)$. Similarly we show that $x_2,x_3 \in B(x_0,r)$ so $B(x_0,r)$ contains the convex set $\{x_1,x_2,x_3\}$. Since the open ball $B(x_0,r)$ is a convex set, we conclude that it also contains the convex hull of $\{x_1,x_2,x_3\}$ which is $\Delta$.