Given a closed smooth curve ${\displaystyle \alpha :[0,L]\to \mathbb {R} ^{3}}$  with unit speed, the velocity ${\displaystyle \gamma ={\dot {\alpha }}:[0,L]\to \mathbb {S} ^{2}}$  is also a closed smooth curve. The total absolute curvature is its length ${\displaystyle l(\gamma )}$ .

The curve ${\displaystyle \gamma }$  does not lie in an open hemisphere. If so, then there is ${\displaystyle v\in \mathbb {S} ^{2}}$  such that ${\displaystyle \gamma \cdot v>0}$ , so ${\displaystyle \textstyle 0=(\alpha (1)-\alpha (0))\cdot v=\int _{0}^{L}\gamma (t)\cdot v\,\mathrm {d} t>0}$ , a contradiction. This also shows that if ${\displaystyle \gamma }$  lies in a closed hemisphere, then ${\displaystyle \gamma \cdot v\equiv 0}$ , so ${\displaystyle \alpha }$  is a plane curve.

Consider a point ${\displaystyle \gamma (T)}$  such that curves ${\displaystyle \gamma ([0,T])}$  and ${\displaystyle \gamma ([T,L])}$  have the same length. By rotating the sphere, we may assume ${\displaystyle \gamma (0)}$  and ${\displaystyle \gamma (T)}$  are symmetric about the axis through the poles. By the previous paragraph, at least one of the two curves ${\displaystyle \gamma ([0,T])}$  and ${\displaystyle \gamma ([T,L])}$  intersects with the equator at some point ${\displaystyle p}$ . We denote this curve by ${\displaystyle \gamma _{0}}$ . Then ${\displaystyle l(\gamma )=2l(\gamma _{0})}$ .

We reflect ${\displaystyle \gamma _{0}}$  across the plane through ${\displaystyle \gamma (0)}$ , ${\displaystyle \gamma (T)}$ , and the north pole, forming a closed curve ${\displaystyle \gamma _{1}}$  containing antipodal points ${\displaystyle \pm p}$ , with length ${\displaystyle l(\gamma _{1})=2l(\gamma _{0})}$ . A curve connecting ${\displaystyle \pm p}$  has length at least ${\displaystyle \pi }$ , which is the length of the great semicircle between ${\displaystyle \pm p}$ . So ${\displaystyle l(\gamma _{1})\geq 2\pi }$ , and if equality holds then ${\displaystyle \gamma _{0}}$  does not cross the equator.

Therefore, ${\displaystyle l(\gamma )=2l(\gamma _{0})=l(\gamma _{1})\geq 2\pi }$ , and if equality holds then ${\displaystyle \gamma }$  lies in a closed hemisphere, and thus ${\displaystyle \alpha }$  is a plane curve.