For a closed, regular space curve of length $\ell$, show that $\int_0^\ell\kappa(s)ds\geq 2\pi$, where $\kappa$ gives curvature Question:
How can one show that the following proposition is true (Only outline of the proof is needed)?
For every closed and regular space curve $c:[a,b] \to \mathbb R^3$ of total length $l$ one has the inequality
$$\int_0^l \kappa (s) ds \ge 2 \pi,$$
with equality if and only if the curve is a convex, simple plane curve$.^1$
Definitions:
A regular parametrized curve is a continuously differentiable immersion $c: I \to \mathbb R^n$, defined on a real interval $I \subseteq \mathbb R$. This means that $\dot{c}=\frac{dc}{dt}\neq0$ holds everywhere. The length of the curve $c$ ($I=[a,b]$) is $$\int_a^b\left \Vert \frac {dc}{dt} \right \Vert dt \,.$$ The function $\kappa:=\left \Vert c'' \right \Vert$ is called the curvature of $c\,.^2$
A (regular) curve $c:[a,b] \to \mathbb R^n$ is called closed, if there is a (regular) curve $\tilde{c}|_{[a,b]}=c$ and $\tilde{c}(t+b-a)=\tilde{c}(t)$ for all $t \in \mathbb R $, where in particular $c(a)=c(b)$ and $c'(a)=c'(b)$. A closed curve $c$ is said to be simply closed, if $c|_{[a,b]}$ is injective$.^3$
A simply closed plane curve is called convex, if the image set of the boundary is a convex subset $C \subset \mathbb R^2 \, .^4$

[1], [2], [3], [4] Wolfgang Kühnel, "Differential Geometry Curves-Surfaces-Manifolds", Second Edition, American Mathematical Society, 2006.
 A: Given the parametrization c of your curve, let p be any point on the unit sphere. Consider the function $g(t)=p \cdot c(t)$. Since the interval $[a,b]$ is closed and bounded, and $g(t)$ is continuous, then it attains a maximum and minimum value  within the interval. It is possible to compute this value by differentiating $g(t)$. Now
\begin{equation}
g'(t)=p \cdot c'(t)
\end{equation} 
Now the points $x \in \mathbb{S}^2$ such that $x \cdot p =0 $ belongs to the great circle of the sphere. The great circle belong to a plane perpendicular to the line connecting the point p to the certer of the sphere. Since $c'$ and the unit tangent vector $T$ are collinear, while p is an arbitrary point on the sphere, we conclude that the tangent indicatrix intersects every great circle on $\mathbb{S}^2$. Since the length of the tangent indicatrix is the total curvature, we have now to prove that the length of the tangent indicatrix is larger of equal to $2 \pi$. To do that you can use the Horn lemma.


*

*Horn lemma
Given a regular curve c on the unit sphere, if c has length less than 2π then c is contained in a hemisphere.

A: In 3 dimensions it should be noted that the sign of the curvature of a smooth space curve is not necessarily well-defined.
To experiment with the total curvature of a space curve, it can be useful to imagine a simple closed polygon P in space — a sequence of line segments [a_0,a_1], [a_1,a_2], ..., [a_(n-1),a_0].
Then this polygon P can be approximated arbitrarily well by smooth curves whose curvature is concentrated in a small neighborhood of the vertices of P. In such a way that the integral of the (absolute) curvature is the sum of all the absolute values of the angles between two consecutive sides of the polygon. 
Each is the absolute angle that the tangent of such a curve would turn when traversing that angle. I.e., the first angle would be cos^(-1) (D), where D is the dot product of 
(a_1-a_0)/||a_1-a_0|| 
and 
(a_2-a_1)/||a_2-a_1||.
A: If $c$ is unit speed, then $T(t)=c'(t)$ is a spherical curve in $\mathbb{S}^2$, which is a closed curve. If ${\rm image}\ (T)$ is in some open hemisphere $H$ s.t. $\partial H$ is a great circle which is orthogonal to some $U\in \mathbb{S}^2$, then $c$ goes to the direction $U$ or $-U$. In any case, $c$ can not return to a point $c(0)$.
Hence ${\rm Image} (T)$ can not be in any open hemisphere, which means that ${\rm length}\ (T)\geq 2\pi$ (cf. Spherical lemma).
